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Q.1. If A and B are points in the plane such that PA/PB = k (constant) for all P on a given circle, then the value of k cannot be equal to ..................... (1982  2 Marks)
Ans. 1
Sol. As P lies on a circle and A and B two points in the plane
such that
Then k can be any real number except 1 as otherwise P will lie on perpendicular bisector of AB which is a line.
Q.2. The points of intersection of the line 4x – 3y – 10 = 0 and the circle x^{2 }+ y^{2} – 2x + 4y – 20 = 0 are .................... and ..................... (1983  2 Marks)
Ans. (4, 2), (–2, –6)
Sol. For point of intersection of line 4x – 3y – 10 = 0 … (1)
and circle x^{2} + y^{2 }– 2x + 4y – 20 = 0 … (2)
Solving (1) and (2), we get
⇒ y^{2} + 4y – 12 = 0 ⇒ y = 2, – 6
⇒ x = 4,– 2
∴ Points are (4, 2) and (– 2, – 6)
Q.3. The lines 3x – 4y + 4 = 0 and 6x – 8y – 7 = 0 are tangents to the same circle. The radius of this circle is ..................... (1984  2 Marks)
Ans. 3/4
Sol. Let 3x – 4y + 4 = 0 be the tangent at point A and 6x – 8y – 7 = 0 be the tangent of point B of the circle.
As the two tangents parallel to each other
∴ AB should be the diameter of circle.
∴ AB = distance between parallel lines
3x – 4y + 4 = 0 and
6x – 8y – 7 = 0 = distance between
6x – 8y + 8 = 0 and 6x – 8y – 7 = 0
∴ radius of circle =
Q.4. Let x^{2} + y^{2 }– 4x – 2y – 11 = 0 be a circle. A pair of tangents from the point (4, 5) with a pair of radii form a quadrilateral of area ..................... (1985  2 Marks)
Ans. 8 sq. units
Sol. KEY CONCEPT :
Length of tangent from a point (x_{1}, y_{1}) to a circle x^{2} + y^{2} + 2gx + 2fy + c = 0 is given by
The equation of circle is, x^{2 }+ y^{2} – 4x – 2y – 11 = 0
It's centre is (2, 1), radius =
length of tangent from the pt. (4, 5) is
∴ Area of quad. ABCD
= 2 (Area of ΔABC)
= 8 sq. units.
Q.5. From the origin chords are drawn to the circle (x – 1)^{2} + y^{2} = 1. The equation of the locus of the midpoints of these chords is ..................... (1985  2 Marks)
Ans. x^{2} + y^{2} – x = 0
Sol. The equation of given circle is (x – 1)^{2} + y^{2} = 1
or x^{2} + y^{2} – 2x = 0 … (1)
KEY CONCEPT : We know that equation of chord of curve S = 0, whose mid point is (x_{1}, y_{1}) is given by T = S_{1} where T is tangent to curve S = 0 at (x_{1}, y_{1}).
∴ If (x_{1}, y_{1}) is the mid point of chord of given circle (1), then equation of chord is
At it passes through origin, we get
Q.6. The equation of the line passing through the points of intersection of the circles 3x^{2} + 3y^{2} – 2x + 12y – 9 = 0 and x^{2 }+ y^{2 }+ 6x + 2y – 15 = 0 is ..................... (1986  2 Marks)
Ans. 10x – 3y – 18 = 0
Sol. The equation of two circles are
… (1)
and x^{2 }+ y^{2} + 6x + 2y – 15 = 0 … (2)
Now we know eq. of common chord of two circles
S_{1 }= 0 and S_{2} = 0 is S_{1} – S_{2 }= 0
10x – 3y – 18 = 0
Q.7. From the point A(0, 3) on the circle x^{2} + 4x + (y – 3)^{2} = 0, a chord AB is drawn and extended to a point M such that AM = 2AB. The equation of the locus of M is .................. (1986  2 Marks)
Ans.
Sol. The equation of circle is, x^{2} + y^{2} + 4x – 6y + 9 = 0 … (1)
AM = 2AB
⇒ AB = BM
Let the coordinates of M be (h, k) Then B is mid pt of AM
As B lies on circle (1),
⇒ h^{2 }+ k^{2} + 8h – 6k + 9 = 0
∴ locus of (h, k) is, x^{2} + y^{2} + 8x – 6y + 9 = 0
Q.8. The area of the triangle formed by the tangents from the point (4, 3) to the the circle x^{2} + y^{2} = 9 and the line joining their points of contact is ..................... (1987  2 Marks)
Ans. 192/25
Sol. From P (4, 3) two tangents PT and PT' are drawn to the circle x^{2 }+ y^{2} = 9 with O (0, 0) as centre and r = 3.
To find the area of ΔPTT'.
Let R be the point of intersection of OP and TT'.
Then we can prove by simple geometry that OP is perpendicular bisector of TT'.
Equation of chord of contact TT' is 4x + 3y = 9 Now, OR = length of the perpendicular from O to TT' is
OT = radius of circle = 3
Again OP =
∴ PR = OP  OR =
Area of the triangle
PTT' = PR x TR
Q.9. If the circle C_{1} : x^{2 }+ y^{2} = 16 intersects anoth er circle C_{2 }of radius 5 in such a manner that common chord is of maximum length and has a slope equal to 3/4, then the coordinates of the centre of C_{2 }are ..................... (1988  2 Marks)
Ans.
Sol. We have C_{1} : x^{2 }+ y^{2} = 16, Centre O_{1} (0, 0) radius = 4. C_{2} is another circle with radius 5, let its centre O_{2} be (h, k).
Now the common chord of circles C_{1 }and C_{2} is of maximum length when chord is diameter of smaller circle C_{1}, and then it passes through centre O_{1} of circle C_{1}. Given that slope of this chord is 3/4.
∴ Equation of AB is,
… (1)
In right ΔAO_{1}O_{2,}
Also distance from (h, k) to (1)
⇒ 3h – 4k ± 15 = 0 … (2)
… (3)
Solving, 3h – 4k + 15 = 0 and 4h + 3k = 0
We get h = – 9/5, k = 12/5
Again solving 3h – 4k – 15 = 0 and 4h + 3k = 0
We get h = 9/5, k = –12/5
Thus the required centre is
Q.10. The area of the triangle formed by the positive xaxis and the normal and the tangent to the circle x^{2} + y^{2 }= 4 at is, ..................... (1989  2 Marks)
Ans.
Sol. Tangent at P (1, ) to the circle x^{2} + y^{2} = 4 is x . 1 + y . = 4
It meets xaxis at A (4, 0),
∴ OA = 4 Also OP = radius of circle = 2
∴ Area of ΔOPA =
sq. units
Q.11. If a circle passes through the points of intersection of the coordinate axes with the lines λx – y + 1 = 0 and x – 2y + 3 = 0, then the value of λ = ................... (1991  2 Marks)
Ans. 2
Sol. The given lines are lx – y + 1 = 0 and x – 2y + 3 = 0 which meet xaxis at and B (– 3, 0) and
yaxis at C (0, 1) and D respectively..
Then we must have, OA × OB = OC × OD
Q.12. The equation of the locus of the midpoints of the circle 4x^{2} + 4y^{2} – 12x + 4y + 1 = 0 that subtend an angle of 2π/ 3 at its centre is ..................... (1993  2 Marks)
Ans. 16x^{2} + 16y^{2} – 48x + 16y + 31 = 0
Sol. The given circle is, 4x^{2 }+ 4y^{2} – 12x + 4y + 1 = 0
or with centre
and
Let M (h, k) be the mid pt. of the chord AB of the given circle, then CM ⊥ AB. ∠ ACB = 120°.
In ΔACM,
and ∠A = 30°
⇒ 16h^{2} + 16k^{2} – 48h + 16k + 31 = 0
∴ locus of (h, k) is 16x^{2} + 16y^{2} – 48x + 16y + 31 = 0
Q.13. The intercept on the line y = x by the circle x^{2} + y^{2 }– 2x = 0 is AB. Equation of the circle with AB as a diameter is ..................... (1996  1 Mark)
Ans. x^{2} + y^{2 }– x – y = 0
Sol. Equation of any circle passin g through the point of intersection of x^{2} + y^{2} – 2x = 0 and y
= x is x^{2} + y^{2} – 2x + l (y – x) = 0
or x^{2} + y^{2} – (2 + l)x + ly = 0
Its centre is
For AB to be the diameter of the required circle, the centre must lie on AB. That is,
Thus, equation of required circle is x^{2} + y^{2} – 2x – y + x = 0 or x^{2} + y^{2 }– x – y = 0
Q.14. For each natural number k, let C_{k} denote the circle with radius k centimetres and centre at the origin. On the circle C_{k}, aparticle moves k centimetres in the counterclockwise direction. After completing its motion on C_{k}, the particle moves to C_{k+1} in the radial direction. The motion of the particle continues in this manner. The particle starts at (1, 0).
If the particle crosses the positive direction of the xaxis for the first time on the circle C_{n} then n = ..................... (1997  2 Marks)
Ans. 7
Sol.
The radius of circle C_{1} is 1 cm, C_{2} is 2 cm and soon.
It starts from A_{1} (1, 0) on C_{1}, moves a distance of 1 cm on C_{1} to come to B_{1}. The angle subtended by A_{1}B_{1} at the centre
will be radians, i.e. 1 radian.
From B1 it moves along radius, OB_{1} and comes to A_{2 }on circle C_{2} of radius 2. From A_{2} it moves on C_{2 }a distance 2 cm and comes to B_{2}. The angle subtended by A_{2}B_{2} is again as before 1 radian. The total angle subtended at the centre is 2 radians. The process continues. In order to cross the xaxis again it must describe 2p radians i.e radians.
Hence it must be moving on circle C_{7}
∴ n = 7
Q.15. The chords of contact of the pair of tangents drawn from each point on the line 2x+y =4 to circle x2+y2 = 1 pass through the point ..................... (1997  2 Marks)
Ans.
Sol. Let (h, k) be any point on the given line
∴ 2h + k = 4 and chord of contact is hx + ky = 1 or hx + (4 – 2h) y = 1
or (4y – 1) + h (x – 2y) = 0
P + l Q = 0.It passes through the intersection of P = 0 and
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