Q.1. Find the equation of the circle whose radius is 5 and which touches the circle x^{2 }+ y^{2} – 2x – 4y – 20 = 0 at the point (5, 5). (1978)
Ans. x^{2} + y^{2} – 18x – 16y + 120 = 0
Sol. The given circle is
x^{2} + y^{2} – 2x – 4y – 20 = 0 whose centre is (1, 2) and radius = 5 Radius of required circle is also 5.
Let its centre be C_{2} (α, β). Both the circles touch each other at P (5, 5).
It is clear from figure that P (5, 5) is the midpoint of C_{1}C_{2}.
Therefore, we should have
and β = 8
∴ Centre of requir ed cir cle is (9, 8) and equation of required circle is
(x – 9)^{2} + (y – 8)^{2} = 5^{2 }
⇒ x^{2} + y^{2} – 18x – 16y + 120 = 0
Q.2. Let A be the centre of the circle x^{2} + y^{2 }– 2x – 4y – 20 = 0.
Suppose that the tangents at the points B(1, 7) and D(4. –2) on the circle meet at the point C. Find the area of the quadrilateral ABCD. (1981  4 Marks)
Ans. 75 sq. units
Sol. The eq. of circle is
x^{2} + y^{2 }– 2x – 4y – 20 = 0
Centre (1, 2), radius =
Using eq. of tangent at (x_{1}, y_{1}) of
x^{2} + y^{2} + 2gx_{1} + 2fy_{1} + c = 0 is
xx_{1} + yy_{1} + g (x + x_{1}) f (y + y_{1}) + c = 0
Eq. of tangent at (1, 7) is x . 1 + y . 7 – (x + 1) – 2 (y + 7) – 20 = 0
⇒ y – 7 = 0 … (1) Similarly eq. of tangent at (4, – 2) is
4x – 2y – (x + 4) –2 (y – 2) – 20 = 0
⇒ 3x – 4y – 20 = 0 … (2)
For pt C, solving (1) and (2),
we get x = 16, y = 7 ∴ C (16, 7).
Now, clearly ar (quad ABCD) = 2 Ar (rt ΔABC)
where AB = radius of circle = 5 and BC = length of tangent from C to circle
∴ ar (quad ABCD) = 5 × 15 = 75 sq. units.
Q.3. Find the equations of the circle passing through (–4, 3) and touching the lines x + y = 2 and x – y = 2. (1982  3 Marks)
Ans.
Sol. Given st. lines are x + y = 2
x – y = 2
As centre lies on ∠ bisector of given equations (lines) which are the lines y = 0 and x = 2.
∴ Centre lies on x axis or x = 2.
But as it passes through (– 4, 3), i.e., II quadrant.
∴ Centre must lie on xaxis Let it be (a, 0) then distance between (a, 0) and (– 4, 3) is = length of ⊥^{ lar} distance from (a, 0) to x + y – 2 = 0
⇒ (a + 4)^{2 }+ (0 – 3)^{2} =
⇒ a^{2} + 20a + 46 = 0 ⇒
∴ Equation of circle is
Q.4. Through a fixed point (h, k) secants are drawn to the circle x^{2} + y^{2} = r^{2}. Show that the locus of the midpoints of the secants intercepted by the circle is x^{2} + y^{2} = hx + ky. (1983  5 Marks)
Ans. Sol. Equation of chord whose mid point is given is
T = S_{1}
[Consider (x_{1}, y_{1}) be mid pt. of AB]
⇒ xx_{1} + yy_{1} – r^{2} = x_{1}^{2} + y_{1}^{2}r^{2}
As it passes through (h, k),
∴ hx_{1} + ky_{1} = x_{1}^{2} +y_{1}^{2 }
∴ locus of (x_{1}, y_{1}) is, x^{2} + y^{2} = hx + ky
Q.5. The abscissa of the two points A and B are the roots of the equation x^{2} + 2ax – b^{2 }= 0 and their ordinates are the roots of the equation x^{2} + 2px – q^{2} = 0. Find the equation and the radius of the circle with AB as diameter. (1984  4 Marks)
Ans.
Sol. Let the two points be A = (α_{1}, β_{1}) and B = (α_{2}, β_{2})
Thus α_{1}, α_{2} are roots of x^{2} + 2αx – b^{2 }= 0
∴ α_{1} + α_{2} = – 2α … (1)
α_{1 }α_{2} = – β_{2} … (2)
β_{1}, β_{2} are roots of x^{2} + 2px – q^{2 }= 0
∴ β_{1} + β_{2} = – 2p … (3)
β_{1}β_{2 }= – q2 … (4)
Now equation of circle with AB as diameter is (x – α_{1}) (x – α_{2}) + (y – β_{1}) (y – β_{2}) = 0
⇒ x^{2}– (α_{1 }+ α_{2})x + α_{1}α_{2} + y^{2}– (β_{1} + β_{2})y + β_{1}β_{2} = 0
⇒ x^{2} + 2αx – β_{2 }+ y^{2} + 2py – q^{2} = 0 [Using eq. (1), (2), (3) and (4)]
⇒ x^{2 }+ y^{2} + 2αx + 2py – b^{2} – q^{2} = 0
Which is the equation of required circle, with its centre (– a, – p) and radius
Q.6. Lines 5x + 12y – 10 = 0 and 5x – 12y – 40 = 0 touch a circle C_{1} of diameter 6. If the centre of C_{1} lies in the first quadrant, find the equation of the circle C_{2} which is concentric with C_{1 }and cuts intercepts of length 8 on these lines. (1986  5 Marks)
Ans. Sol. Let equation of tangent PAB be 5x + 12y – 10 = 0 and that of
PXY be 5x – 12y – 40 = 0
Now let centre of circles C_{1} and C_{2} be C (h, k).
Let CM ⊥ PAB then CM = radius of C_{1} = 3
Also C_{2} makes an intercept of length 8 units on PAB ⇒ AM = 4
Then in ΔAMC, we get
∴ Radius of C_{2} is = 5 units
Also, as 5x + 12y – 10 = 0 … (1)
and 5x – 12y – 40 = 0 … (2)
are tangents to C_{1}, length of perpendicular from C to AB = 3 units
⇒ 5h + 12k – 49 = 0 … (i)
or5h + 12k + 29 = 0 … (ii)
⇒ 5h – 12k – 79 = 0 … (iii)
or5h – 12k – 1 = 0 … (iv)
As C lies in first quadrant
∴ h, k are + ve
∴ Eq. (ii) is not possible.
Solving (i) and (iii), we get h = 64/5, k = – 5/4
This is also not possible.
Now solving (i) and (iv), we get h = 5, k = 2.
Thus centre for C_{2 }is (5, 2) and radius 5.
Hence, equation of C_{2} is (x – 5)^{2 }+ (y – 2)^{2} = 5^{2}
⇒ x^{2} + y^{2 }– 10x – 4y + 4 = 0
Q.7. Let a given line L_{1} intersects the x and y axes at P and Q, respectively. Let another line L_{2}, perpendicular to L1, cut the x and y axes at R and S, respectively. Show that the locus of the point of intersection of the lines PS and QR is a circle passing through the origin. (1987  3 Marks)
Ans.
Sol. Let the equation of L_{1} be x cosα + y sin a = p_{1}
Then any line perpendicular to L_{1} is x sin a – y cosα = p_{2}, where p_{2} is a variable.
Then L_{1} meets xaxis at P (p_{1} secα, 0) and yaxis at Q (0, p_{1} cosecα).
Similarly L_{2 }meets xaxis at R (p_{2} cosecα, 0) and yaxis at S (0, – p_{2} secα).
Now equation of PS is,
Similarly, equation of QR is,
… (2)
Locus of point of intersection of PS and QR can be obtained by eliminating the variable p_{2} from (1) and (2)
i.e.
[Substituting the value of from (1) in (2)]
⇒ (x – p_{1} secα) x + y^{2} = p_{1}y cosecα
⇒ x^{2} + y^{2} – p_{1} x secα – p_{1}y cosecα = 0
which is a circle through origin.
Q.8. The circle x^{2} + y^{2} – 4x– 4y + 4 = 0 is inscribed in a triangle which has two of its sides along the coordinate axes. The locus of the circumcentre of the triangle is x + y – xy + k(x^{2 }+ y^{2})^{1/2} = 0. Find k. (1987  4 Marks)
Ans. Sol. The given circle is x^{2} + y^{2} – 4x – 4y + 4 = 0.
This can be rewritten as (x – 2)^{2} + (y – 2)^{2} = 4
which has centre C (2, 2) and radius 2.
Let the eq. of third side AB of ΔOAB issuch that
A (a, 0) and B (0, b)
Length of perpendicular form (2, 2) on AB = radius = CM = 2
Since (2, 2) and origin lie on same side of AB
… (1)
Since ∠AOB = π/2.
Hence, AB is the diameter of the circle passing through ΔOAB, mid point of AB is the centre of the circle i.e. (a/2, b/2)
Let centre be (h, k)
then a = 2h, b = 2k.
Substituting the values of a and b in (1), we get
∴ Locus of M (h, k) is,
… (2)
Comparing it with given equation of locus of circumcentre of Δ i.e.
…(3)
We get, k = 1
Q. 9. If i = 1, 2, 3, 4 are four distinct points on a circle, then show that m_{1}m_{2} m_{3}m_{4} = 1 (1989  2 Marks)
Ans. Sol. Given that i = 1, 2, 3, 4 are four distinct points on a circle.
Let the equation of circle be x^{2} + y^{2} + 2gx + 2fy + c = 0
As the point lies on it, therefore, we have
⇒ m^{4 }+ 2gm^{3} + cm^{2} + 2fm + 1 = 0
Since m_{1}, m_{2}, m_{3}, m_{4} are roots of this equation, therefore product of roots = 1
⇒ m_{1}m_{2}m_{3}m_{4} = 1
Q.10. A circle touches the line y = x at a point P such th at , where O is the origin. The circle contains the point (– 10, 2) in its interior and the length of its chord on the line x + y = 0 is . Determine the equation of the circle. (1990  5 Marks)
Ans. x^{2} + y^{2 }+ 18x – 2y + 32 = 0
Sol. Let AB be the length of ch ord intercepted by circle on y + x = 0
Let CM be perpendicular to AB from centre C (h, k).
Also y – x = 0 and y + x = 0 are perpendicular to each other.
∴ OPCM is rectangle.
∴ CM = OP =
Let r be the radius of cirlce.
∴ In ΔCAM, AC^{2 }= AM^{2} + MC^{2}
Again y = x is tangent to the circle at P
∴ CP = r
(1)
Also CM =
… (2)
Solving four sets of eq’s given by (1) and (2), we get the possible centres as
(9, – 1), (1, – 9), (– 1, 9), (– 9, 1)
∴ Possible circles are (x – 9)^{2} + (y + 1)^{2} – 50
= 0 (x – 1)^{2 }+ (y + 9)^{2 }– 50
= 0 (x + 1)^{2 }+ (y – 9)^{2} – 50
= 0 (x + 9)^{2} + (y – 1)^{2 }– 50 = 0
But the pt (– 10, 2) lies inside the circle.
∴ S_{1} < 0
which is satisfied only for (x + 9)^{2} + (y – 1)^{2} – 50 = 0
∴ The required eq. of circle is x^{2} + y^{2 }+ 18x – 2y + 32 = 0
Q.11. Two circles, each of radius 5 units, touch each other at (1, 2).
If the equation of their common tangent is 4x + 3y = 10, find the equation of the circles. (1991  4 Marks)
Ans. Sol. Let t be the common tangent given by 4x + 3y = 10 … (1)
Common pt of contact being P (1, 2)
Let A and B be the centres of the circles, required.
Clearly, AB is the line perpendicular to t and passing through P (1, 2).
Therefore eq. of AB is
For pt A, r = – 5 and for pt B, r = 5, we get
⇒ For pt A x = – 4 + 1, y = – 3 + 2
and For pt B x = 4 + 1, y = 3 + 2
∴ A (– 3, – 1) B (5, 5).
∴ Eq.’s of required circles are (x + 3)^{2} + (y + 1)^{2} = 5^{2 }
and (x – 5)^{2} + (y – 5)^{2 }= 5^{2}
Q.12. Let a circle be given by 2x(x – a) + y(2y – b) = 0, (a ≠ 0, b ≠ 0).
Find the condition on a and b if two chords, each bisected by the x axis, can be drawn to the circle from (1992  6 Marks)
Ans. Sol. The given circle is 2x (x – a) + y (2y – b) = 0 (a, b ≠ 0)
⇒ 2x^{2} + 2y^{2 }– 2ax – by = 0 ....(1)
Let us consider the chord of this circle which passes through
the pt and whose mid pt. lies on xaxis.
Let (h, 0) be the mid point of the chord, then eq. of chord can be obtained by T = S_{1}
i.e., 2xh + 2y.0 – a (x + h) – (y + 0) = 2h^{2} – 2ah
⇒ (2h – a) x – y + ah – 2h^{2} = 0
This chord passes through therefore
⇒ 8h^{2} – 12ah + (4a^{2} + b^{2}) = 0
As given in question, two such chords are there, so we should have two real and distinct values of h from the above quadratic in h, for which
D > 0
⇒ (12a)^{2} – 4 × 8 × (4 a^{2} + b^{2}) > 0
⇒ a^{2} > 2b^{2}
Q.13. Consider a family of circles passing through two fixed points A (3,7) and B (6, 5). Show that the chords in which the circle x^{2} + y^{2} – 4x – 6y – 3 = 0 cuts the members of the family are concurrent at a point. Find the coordinate of this point. (1993  5 Marks)
Ans.
Sol. Let the family of circles, passing through A (3, 7) and B (6, 5), be
x^{2} + y^{2} + 2gx + 2fy + c = 0
As it passes through (3, 7)
∴ 9 + 49 + 6g + 14f + c = 0 or,6g + 14f + c + 58 = 0 … (1)
As it passes through (6, 5)
∴ 36 + 25 + 12g + 10f + c = 0
12g + 10f + c + 61 = 0 … (2)
(2) – (1) gives,
6g – 4f + 3 = 0
Substituting the value of g in equation (1),
we get 4f – 3 + 14f + c + 58 = 0
⇒ 18f + 55 + c = 0 ⇒ c = – 18f – 55
Thus the family is
x + 2fy – (18f + 55) = 0
Members of this family are cut by the circle x^{2} + y^{2} – 4x – 6y – 3 = 0
∴ Equation of family of chords of intersection of above two circles is S_{1} – S_{2} = 0
x + (2f + 6) y – 18f + 52) = 0
which can be written as
which represents the family of lines passing through the pt. of intersection of the lines
3x + 6y – 52 = 0 and 4x + 6y – 54 = 0
Solving which we get x = 2 and y = 23/3.
Thus the required pt. of intersection is
Q.14. Find the coordinates of the point at which the circles x^{2} + y^{2} – 4x – 2y = –4 and x^{2} + y^{2} – 12x – 8y = – 36 touch each other. Also find equations common tangents touching the circles in the distinct points. (1993  5 Marks)
Ans. Sol. The given circles are
x^{2} + y^{2} – 4x – 2y = – 4 and
x^{2} + y^{2} – 12x – 8y = – 36
i.e., x^{2} + y^{2} – 4x – 2y + 4 = 0 … (1)
x^{2} + y^{2} – 12x – 8y + 36 = 0 .... (2)
with centres C_{1}(2, 1) and C_{2} (6, 4) and radii 1 and 4 respectively.
Also C_{1}C_{2} = 5 As r_{1} + r_{2} = C_{1}C_{2}
⇒ Two circles touch each other externally, at P.
Clearly, P divides C_{1}C_{2 }in the ratio 1 : 4
∴ Coordinates of P are
Let AB and CD be two common tangents of given circles, meeting each other at T. Then T divides C_{1}C_{2} externally in the ratio 1 : 4.
KEY CONCEPT : [As the direct common tangents of two circles pass through a pt. which divides the line segment joining the centres of two circles externally in the ratio of their radii.]
Hence,
Let m be the slope of the tangent, then equation of tangent through (2/3, 0) is
Now, length of perpendicular from (2, 1), to the above tangent is radius of the circle
⇒ (3 – 4m)^{2 }= 9(m^{2} + 1) ⇒ 9 – 24m + 16m^{2} = 9m^{2} + 9
⇒ 7m^{2} – 24m = 0 ⇒ m = 0,
Thus the equations of the tangents are y = 0 and 7y – 24x + 16 = 0.
Q.15. Find the intervals of values of a for which the line y + x = 0 bisects two chords drawn from a point to the circle 2x^{2} + 2y^{2 } (1996  5 Marks)
Ans.
Sol. Let the given point be
and the equation of the circle
becomes x^{2} + y^{2 }– px –
Since the chord is bisected by the line x + y = 0, its midpoint can be chosen as (k, – k). Hence the equation of the chord by T = S_{1} is
It passes through
or … (1)
Put … (2)
Hence, from (1) by the help of (2), we get
… (3)
Since, there are two chords which are bisected by x + y = 0, we must have two real values of k from (3)
∴ Δ > 0
or 18a^{2} – 8 (1 + 2a^{2}) > 0
or,a^{2} – 4 > 0
or,(a + 2) (a – 2) > 0
∴ a < – 2 or > 2
∴ a ∈ (– ∞, – 2) ∪ (2, ∞)
or a ∈ ] – ∞, –2[ ∪ ] 2, ∞ [
Q.16. A circle passes through three points A, B and C with the line segment AC as its diameter. A line passing through A intersects the chord BC at a point D inside the circle. If angles DAB and CAB are α and β respectively and the distance between the point A and the mid point of the line segment DC is d, prove that the area of the circle is
(1996  5 Marks)
Ans.
Sol. Let r be the radius of circle, then AC = 2r Since, AC is the diameter
∴ In ΔABC BC = 2r sinβ, AB = 2r cosβ
In rt ∠ed ΔABC BD = AB tanα = 2r cosβ tanα
AD = AB secα = 2r cosβ secα
∴ DC = BC – BD = 2r sinβ – 2r cosβ tanα
Now since E is the mid point of DC
⇒ DE = r sinβ – r cosβ tanα Now in ΔADC, AE is the median
∴ 2 (AE^{2} + DE^{2}) = AD^{2} + AC^{2}
⇒ 2 [d^{2} + r^{2} (sinβ – cosβ tanα)^{2}]
= 4r^{2} cos^{2} b sec^{2} a + 4r^{2}
⇒ Area of circle,
Q.17. Let C be any circle with centre (0, ) . Prove that at the most two rational points can be there on C. (A rational point is a point both of whose coordinates are rational numbers.) (1997  5 Marks)
Ans.
Sol. Given C is the circle with centre at (0, 2 ) and radius r (say) then
… (1)
The only rational value which y can have is 0.
Suppose the possible value of x for which y is 0 is x_{1}. Certainly – x_{1} will also give the value of y as 0 (from (1)).
Thus, at the most, there are two rational pts which satisfy the eq^{n} of C.
Q.18. C_{1} and C_{2} are two concentric circles, the radius of C_{2} being twice that of C1. From a point P on C_{2}, tangents PA and PB are drawn to C_{1}. Prove that the centroid of the triangle PAB lies on C_{1}. (1998  8 Marks)
Ans.
Sol. Let P (h, k) be on C_{2}
∴ h^{2 }+ k^{2} = 4r^{2}
Chord of contact of P w.r.t. C_{1} is hx + ky = r^{2 }
It intersects C_{1}, x^{2} + y^{2} = a^{2} in A and B.
Eliminating y, we get,
If (x, y) be the centroid of ΔPAB, then
or h = 2x and similarly k = 2y
Putting in (1) we get 4x^{2 }+ 4y^{2} = 4r^{2 }
∴ Locus is x^{2 }+ y^{2} = r^{2} i.e., C_{1}
Q.19. Let T_{1}, T_{2} be two tangents drawn from (–2, 0) onto the circle C : x^{2} + y^{2} = 1. Determine the circles touching C and having T_{1}, T_{2} as their pair of tangents. Further, find the equations of all possible common tangents to these circles, when taken two at a time. (1999  10 Marks)
Ans.
Sol. The given circle is x^{2} + y^{2} = 1 … (1)
Centre O (0, 0) radius = 1
Let T_{1} and T_{2} be the tangents drawn from (– 2, 0) to the circle (1).
Let m be the slope of tangent then equations of tangents are
y – 0 = m (x + 2) or, mx – y + 2m = 0 … (2)
As it is tangent to circle (1)
length of ⊥ lar from (0, 0) to (2) = radius of (1)
4m^{2} = m^{2 }+ 1 ⇒ m
∴ Th e two tangen ts are and
Now any other circle touching (1)
and T_{1}, T_{2} is such that its centre lies on xaxis.
Let (h, 0) be the centre of such circle, then from fig.
OC_{1} = OA + AC_{1} ⇒  h  = 1 +  AC_{1} 
But AC_{1 }= ⊥ lar distance of (h, 0) to tangent
Squaring,
⇒ 4h^{2} ± 8h + 4 = h^{2} + 4h + 4
‘ + ’ ⇒ 3h^{2} = – 4h ⇒ h = – 4/3
‘–’ ⇒ 3h^{2} = 12h ⇒ h = 4
Thus centres of circles are (4, 0),
∴ Radius of circle with centre (4, 0) is = 4 – 1 = 3 and radius of circle with centre
∴ The two possible circles are (x – 4)^{2} + y^{2 }= 3^{2} … (3)
… (4)
Now, common tangents of (1) and (3). Since (1) and (3) are two touching circles they have three common tangents
T_{1}, T_{2} and x = 1 (clear from fig.)
Similarly common tangents of (1) and (4) are T_{1}, T_{2} and x = – 1.
For the circles (3) and (4) there will be four common tangents of which two are direct common tangents .
XY and x' y' and two are indirect common tangents. Let us find two common indirect tangents. We know that In two similar Δ’s C_{1}XN and C_{2}YN
divides C_{1}C_{2} in the ratio 9 : 1.
Clearly N lies on xaxis.
Any line through N is
or 5mx – 5y + 4m = 0
If it is tangent to (3) then
⇒ 64m^{2} = 25m^{2 }+ 25
⇒ 39m^{2} = 25 ⇒
∴ Required tangents are
Q.20. Let 2x^{2} + y^{2}  3xy = 0 be the equation of a pair of tangents drawn from the origin O to a circle of radius 3 with centre in the first quadrant. If A is one of the points of contact, find the length of OA. (2001  5 Marks)
Ans. Sol. The equation 2x^{2} – 3xy + y^{2} = 0 represents pair of tangents OA and OA'.
Let angle between these to tangents be 2θ.
Then,
As θ is acute
Now we know that line joining the pt through which tangents are drawn to the centre bisects the angle between the tangents,
∴∠AOC = ∠A'OAC = θ
In ΔAOC,
Q.21. Let C_{1} and C_{2} be two circles with C_{2} lying inside C_{1}. A circle C lying inside C_{1} touches C_{1} internally and C_{2} externally.
Identify the locus of the centre of C. (2001  5 Marks)
Ans. Sol. Let equation of C_{1} be x^{2 }+ y^{2} = r_{1}^{2} and of C_{2} be
(x – a)^{2} + (y – b)^{2 }= r_{2}^{2}
Let centre of C be (h, k) and radius be r, then by the given conditions.
and
Required locus is
which represents an ellipse whose foci are at (a, b) and (0, 0).
[∵ PS + PS' = constant ⇒ locus of P is an ellipse with foci at S and S']
Q.22. For the circle x^{2} + y^{2} = r^{2}, find the value of r for which the area enclosed by the tangents drawn from the point P (6, 8) to the circle and the chord of contact is maximum. (2003  2 Marks)
Ans.
Sol. The given circle is x^{2} + y^{2} = r^{2}
From pt. (6, 8) tangents are drawn to this circle.
Then length of tangent
Also equation of chord of contact LM is
6x + 8y – r^{2} = 0
PN = length of ⊥ lar from P to LM
Now in rt. ΔPLN, LN^{2} = PL^{2} – PN^{2}
(∵ LM = 2 LN)
∴ Area of ΔPLM =x L M x PN
For max value of area, we should have
⇒ r = 10 or r = 5
But r = 10 gives length of tangent PL = 0
∴ r ≠ 10. Hence, r = 5
Q.23. Find the equation of circle touching the line 2x + 3y + 1 = 0 at (1, – 1) and cutting orthogonally the circle having line segment joining (0, 3) and (–2, –1) as diameter. (2004  4 Marks)
Ans. Sol. We are given that line 2x + 3y + 1 = 0 touches a circle S = 0 at (1, – 1).
So, eq^{n} of this circle can be given by (x – 1)^{2} + (y + 1)^{2} + λ(2x + 3y + 1) = 0.
[Note : (x – 1)^{2} + (y + 1)^{2} = 0 represents a pt. circle with centre at (1, –1)].
or x^{2} + y^{2} + 2x (λ – 1) + y (3λ + 2) + (λ + 2) = 0 …(1)
But given that this circle is orthogonal to the circle, the extremities of whose diameter are (0, 3) and (– 2, – 1)
i.e. x (x + 2) + (y – 3) (y + 1) = 0
x^{2} + y^{2} + 2x – 2y – 3 = 0 .......... (2)
Applying the condition of orthogonality for (1) and (2), we
[2g_{1}g_{2} + 2f_{1}f_{2} = c_{1} + c_{2}]
⇒ 2λ – 2 – 3λ – 2 = λ – 1
Substituting this value of λ in eq^{n }(1) we get the required circle as
or,2x^{2} + 2y^{2} – 10x – 5y + 1 = 0
Q.24. Circles with radii 3, 4 and 5 touch each other externally. If P is the point of intersection of tangents to these circles at their points of contact, find the distance of P from the points of contact. (2005  2 Marks)
Ans. Sol. Given these circles with centres at C_{1}, C_{2} and C_{3} and with radii 3, 4 and 5 respectively, The three circles touch each other externally as shown in the figure.
P is the point of intersection of the three tangents drawn at the pts of contacts, L, M and N. Since lengths of tangents to a circle from a point are equal, we get
PL = PM = PN
Also PL ⊥ C_{1}C_{2} , PM ⊥ C_{2}C_{3}, PN⊥ C_{1}C_{3}
(Q tangent is perpendicular to the radius at pt. of contact)
Clearly P is the incentre of ΔC_{1}C_{2}C_{3} and its distance from pt.of contact i.e.,
PL is the radius of incircle of ΔC_{1}C_{2}C_{3}.
In ΔC_{1}C_{2}C_{3} sides are a = 3 + 4 = 7, b = 4 + 5 = 9, c = 5 + 3 = 8
447 docs930 tests


Explore Courses for JEE exam
