JEE Advanced (Subjective Type Questions): Circle | Chapter-wise Tests for JEE Main & Advanced PDF Download

Q.1. Find the equation of the circle whose radius is 5 and which touches the circle x² + y² – 2x – 4y – 20 = 0 at the point (5, 5). (1978)

Ans. x² + y² – 18x – 16y + 120 = 0

Sol.  The given circle is 

x² + y² – 2x – 4y – 20 = 0 whose centre is (1, 2) and radius = 5 Radius of required circle is also 5.
Let its centre be C₂ (α, β). Both the circles touch each other at P (5, 5).

It is clear from figure that P (5, 5) is the mid-point of C₁C₂.
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 = 5² 

⇒ x² + y² – 18x – 16y + 120 = 0

Q.2. Let A be the centre of the circle x² + y² – 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
x² + y² – 2x – 4y – 20 = 0
Centre (1, 2), radius =

Using eq. of tangent at (x₁, y₁) of
x² + y² + 2gx₁ + 2fy₁ + c = 0 is
xx₁ + yy₁ + g (x + x₁) f (y + y₁) + 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 x-axis 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 = 

⇒ a² + 20a + 46 = 0  ⇒  

∴ Equation of circle is

Q.4. Through a fixed point (h, k) secants are drawn to the circle x² + y² = r². Show that the locus of the mid-points of the secants intercepted by the circle is x² + y² = hx + ky. (1983 - 5 Marks)

Ans. Sol.  Equation of chord whose mid point is given is

T = S₁

[Consider (x₁, y₁) be mid pt. of AB]

⇒ xx₁ + yy₁ – r² = x₁² + y₁²-r²

As it passes through (h, k),

∴ hx₁ + ky₁ = x₁² +y₁² 

∴ locus of (x₁, y₁) is, x² + y² = hx + ky

Q.5. The abscissa of the two points A and B are the roots of the equation x² + 2ax – b² = 0 and their ordinates are the roots of the equation x² + 2px – q² = 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 = (α₁, β₁)  and B = (α₂, β₂)
Thus α₁, α₂ are roots of x² + 2αx – b² = 0
∴  α₁ + α₂ = – 2α … (1)
α₁ α₂ = – β₂ … (2)
β₁, β₂ are roots of x² + 2px – q² = 0
∴  β₁ + β₂ = – 2p … (3)    
β₁β₂ = – q2 … (4)
Now equation of circle with AB as diameter is (x – α₁) (x – α₂) + (y – β₁) (y – β₂) = 0
⇒ x²– (α₁ + α₂)x + α₁α₂ + y²– (β₁ + β₂)y + β₁β₂ = 0
⇒ x² + 2αx – β₂ + y² + 2py – q² = 0 [Using eq. (1), (2), (3) and (4)]
⇒ x² + y² + 2αx + 2py – b² – q² = 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 C₁ of diameter 6. If the centre of C₁ lies in the first quadrant, find the equation of the circle C₂ which is concentric with C₁ 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 C₁ and C₂ be C (h, k).
Let CM ⊥ PAB  then CM = radius of C₁ = 3
Also C₂  makes an intercept of length 8 units on PAB ⇒ AM = 4

Then in ΔAMC, we get

∴ Radius of C₂ is  = 5 units
Also,  as 5x + 12y – 10 = 0 … (1)
and 5x – 12y – 40 = 0 … (2)
are tangents to C₁, 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 C₂ is (5, 2) and radius 5.
Hence, equation  of C₂ is (x – 5)² + (y – 2)² = 5²
⇒ x² + y² – 10x – 4y + 4 = 0

Q.7. Let a given line L₁ intersects the x and y axes at P and Q, respectively. Let another line L₂, 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 L₁ be x cosα + y sin a = p₁
Then any line perpendicular to L₁ is x sin a  – y cosα = p₂, where p₂ is a variable.
Then L₁ meets x-axis at P (p₁ secα, 0) and y-axis at Q (0, p₁ cosecα).
Similarly L₂ meets x-axis at R (p₂ cosecα, 0) and y-axis at S (0, – p₂ 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 p₂ from (1) and (2)

i.e.

[Substituting the value of   from (1) in (2)]

⇒ (x – p₁ secα) x + y² = p₁y cosecα
⇒ x² + y² – p₁ x secα – p₁y cosecα = 0
which is a circle through origin.

Q.8. The circle x² + y² – 4x– 4y + 4 = 0 is inscribed in a triangle which has two of its sides along the co-ordinate axes. The locus of the circumcentre of the triangle is x + y – xy + k(x² + y²)^1/2 = 0. Find k. (1987 - 4 Marks)

Ans. Sol. The given circle is x² + y² – 4x – 4y + 4 = 0.
This can be re-written 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 m₁m₂ m₃m₄ = 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 x² + y² + 2gx + 2fy + c = 0

As the point  lies on it, therefore, we have

⇒ m⁴ + 2gm³ + cm² + 2fm + 1 = 0

Since m₁, m₂, m₃, m₄ are roots of this equation, therefore product of roots = 1

⇒ m₁m₂m₃m₄ = 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. x² + y² + 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, AC² = AM² + MC²

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.
∴ S₁ < 0
which is satisfied only for (x + 9)² + (y – 1)² – 50 = 0
∴ The required eq. of circle is x² + y² + 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 = 5² 
and (x – 5)² + (y – 5)² = 5²

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)
⇒ 2x² + 2y² – 2ax – by = 0 ....(1)
Let us consider the chord of this circle which passes through

the pt  and whose mid pt. lies on x-axis.

Let (h, 0) be the mid point of the chord, then eq. of chord can be obtained by T =  S₁

i.e., 2xh + 2y.0 – a (x + h) –  (y + 0) = 2h² – 2ah

⇒ (2h – a) x –  y + ah – 2h² = 0

This chord passes through therefore

⇒ 8h² – 12ah + (4a² + b²) = 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 a² + b²) > 0
⇒ a² > 2b²

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 x² + y² – 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
x² + y² + 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 x² + y² – 4x – 6y – 3 = 0
∴ Equation of family of chords of intersection of above two circles is S₁ – S₂ = 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 x² + y² – 4x – 2y = –4 and x² + y² – 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
x² + y² – 4x – 2y = – 4 and
 x² + y² – 12x – 8y = – 36
i.e.,  x² + y² – 4x – 2y + 4 = 0 …  (1)
x² + y² – 12x – 8y + 36 = 0 ....  (2)
with centres C₁(2, 1) and C₂ (6, 4) and radii 1 and 4 respectively.
Also C₁C₂ = 5 As r₁ + r₂ = C₁C₂
⇒ Two circles touch each other externally, at P.

Clearly, P divides C₁C₂ in the ratio 1 : 4

∴ Co-ordinates of P are

Let AB and CD be two common tangents of given circles, meeting each other at T. Then T divides C₁C₂ 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(m² + 1) ⇒ 9 – 24m + 16m² = 9m² + 9

⇒ 7m² – 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 2x² + 2y² -              (1996 - 5 Marks) 

Ans. 

Sol.  Let the given point be

 and the equation of the circle

becomes x² + y² – px –

Since the chord is bisected by the  line x + y = 0, its mid-point can be chosen as (k, – k). Hence the equation of the chord by T = S₁ 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 18a² – 8 (1 + 2a²) > 0
or,a² – 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 (AE² + DE²) = AD² + AC²
⇒ 2 [d² + r² (sinβ – cosβ tanα)²]
= 4r² cos² b sec² a + 4r²

⇒ 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 x₁. Certainly – x₁ will also give the value of y as 0 (from (1)).
Thus, at the most, there are two rational pts which satisfy the eqⁿ of C.

Q.18. C₁ and C₂ are two concentric circles, the radius of C₂ being twice that of C1. From a point P on C₂, tangents PA and PB are drawn to C₁. Prove that the centroid of the triangle PAB lies on C₁. (1998 - 8 Marks) 

Ans. 

Sol. Let P (h, k) be on C₂
∴ h² + k² = 4r²
Chord of contact of P w.r.t. C₁ is hx + ky = r² 
It intersects C₁, x² + y² = a² 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 4x² + 4y² = 4r² 
∴ Locus is x² + y² = r²  i.e., C₁

Q.19. Let T₁, T₂ be two tangents drawn from (–2, 0) onto the circle C : x² + y² = 1. Determine the circles touching C and having T₁, T₂ 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 x² + y² = 1 … (1)
Centre O (0, 0) radius = 1
Let T₁ and T₂ 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)

4m² = m² + 1  ⇒ m 

∴ Th e two tangen ts are and

Now any other circle touching (1)
and T₁, T₂ is such that its centre lies on x-axis.
Let (h, 0) be the centre of such circle, then from fig.
OC₁ = OA + AC₁ ⇒ | h | = 1 + | AC₁ |
But AC₁ = ⊥ lar distance of (h, 0) to tangent

Squaring,

⇒ 4h² ± 8h + 4 = h² + 4h + 4
‘ + ’ ⇒ 3h² = – 4h ⇒ h = – 4/3
‘–’ ⇒ 3h² = 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 + y² = 3² … (3)

 … (4)
Now, common tangents of (1) and (3). Since (1) and (3) are two touching circles they have three common tangents
T₁, T₂ and x = 1 (clear from fig.)
Similarly common tangents of (1) and (4) are T₁, T₂ 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 C₁XN and C₂YN

divides C₁C₂ in the ratio 9 : 1.

Clearly N lies on x-axis.

Any line through N is

or    5mx – 5y + 4m = 0

If it is tangent to (3) then

     ⇒  64m² = 25m² + 25

⇒ 39m² = 25 ⇒

∴ Required tangents are

Q.20. Let 2x² + y² - 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 2x² – 3xy + y² = 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 C₁ and C₂ be two circles with C₂ lying inside C₁. A circle C lying inside C₁ touches C₁ internally and C₂ externally.
 Identify the locus of the centre of C. (2001 - 5 Marks) 

Ans. Sol.  Let equation of C₁ be x² + y² = r₁² and of C₂ be
(x – a)² + (y – b)² = r₂²

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 x² + y² = r², 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 x² + y² = r²
From pt. (6, 8) tangents are drawn to this circle.

Then length of tangent

Also equation of chord of contact LM is
6x + 8y – r² = 0
PN = length of ⊥ lar from P to LM

Now in rt. ΔPLN, LN² = PL² – PN²

(∵ 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, eqⁿ 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  x² + y² + 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
x² + y² + 2x – 2y – 3 = 0 .......... (2)
Applying the condition of orthogonality for (1) and (2), we

                                          [2g₁g₂ + 2f₁f₂ = c₁ + c₂]

⇒ 2λ – 2 – 3λ – 2 = λ – 1

Substituting this value of λ in eqⁿ (1) we get the required circle as

or,2x² + 2y² – 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 C₁, C₂ and C₃ 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 ⊥ C₁C₂ , PM ⊥ C₂C₃, PN⊥ C₁C₃
(Q  tangent is perpendicular to the radius at pt. of contact)
Clearly P is the incentre of ΔC₁C₂C₃ and its distance from pt.of contact i.e.,
PL is the radius of incircle of ΔC₁C₂C₃.
In ΔC₁C₂C₃ sides are a = 3 + 4 = 7, b = 4 + 5 = 9, c = 5 + 3 = 8