Q: In the given figure, the equation of the larger circle is x^ 2 +y^ 2 +4 y-5=0 and the distance between centres is 4 . Then the equation of smaller circle is

We have x^ 2 +y^ 2 +4 y-5=0. Its centre is C _ 1 (0,-2) r _ 1 = 4+5 =3. Let C _ 2 ( ~h , k ) be the centre of the smaller circle and its radius r _ 2 . Then C _ 1 C _ 2 =4 h ^ 2 +( k +2)^ 2 =3+ r _ 2 =4 r _ 2 =1 But k = r _ 2 =1 [it touches x -axis From eq (1), 4= h ^ 2 +(1+2)^ 2 16= h ^ 2 +9 h ^ 2 =7 h = 7 Since h >0 h = 7 Hence, required circle is (x- 7 )^ 2 +(y-1)^ 2 =1

Question 1

The locus of the mid-point of a chord of the circle x^2+y^2=4, which subtends a right angle at the origin is

Accepted Answer

According to question, O C= h^2+k^2 In O C B, aligned & 45^ = h^2+k^2 2 \ & h^2+k^2 2 = 1 2 \ & h^2+k^2=2 aligned Replacing h x and k y Locus x^2+y^2=2

Question 2

If p and q be the longest and the shortest distance respectively of the point (-7,2) from any point ( , ) on the curve whose equation is x^2+y^2-10 x-14 y-51=0, then G.M. of p

Accepted Answer

The centre C of the circle =(5,7) and the radius = 5^2+7^2+51 =5 5 PC = 12^2+5^2 =13 q=P A=13-5 5 and p=P B=13+5 3 G.M. of p and q = p q = (13-5 5 )(13+5 5 ) = 169-125 =2 11 .

Question 3

From a point A (0,3) on the circle (x+2)^2 +(y-3)^2=4, a chord A B is drawn and it is extended to a point Q such that A Q=2 A B. Then the locus of Q is

Accepted Answer

Given equation of circle (x+2)^2+(y-3)^2=4 Let the coordinates of Q is (h, k). Coordinate of B which is midpoint of AQ because AQ =2 AB . Then, B = ( 0+h 2 , k+3 2 ) ( h 2 , k+3 2 ) Point B also satisfy the equation of circle. (x+2)^2+(y-3)^2=4 ( h 2 +2 )^2+ ( k+3 2 -3 )^2=4 (h+4)^2 4 + (k-3)^2 4 =4 (h+4)^2+(k-3)^2=16 Replace (h, k) by (x, y), then, the required equation is (x+4)^2+(y-3)^2=16

Question 4

The equation of the circle with centre (0,2) and radius 2 is x^2+y^2-m y=0. The value of m is

Accepted Answer

Here h=0, k=2 and r=2. Therefore, the required equation of the circle is : (x-0)^2+(y-2)^2=(2)^2 or x^2+y^2-4 y+4=4 or x^2+y^2-4 y=0

Question 5

A circle touches both the y-axis and the line x+ y=0. Then the locus of its center is

Accepted Answer

Let (h, k) is centre of circle | h-k 2 |=|h|, k^2-h^2+2 h k=0

Question 6

For each parabola y=x^2+p x+q, meeting coordinate axes at 3-distinct points, if circles are drawn through these points, then the family of circles must pass through

Accepted Answer

Suppose the parabola y=x^2+p x+q cuts X-axis at A( , 0) and B( , 0). Then, , are roots of the equation x^2+p x+q=0 + =-p and =q The parabola y=x^2+p x+q cuts Y-axis at (0, q). Let the equation of the circle passing through A, B and C be aligned & x^2+y^2+2 g x+2 f y+c =0 &...(i)\ & ^2+2 g +c =0 &...(ii)\ & ^2+2 g +c =0 &...(iii)\ and & q^2+2 f q+c =0 &...(iv) aligned Subtracting Eq. (iii) from Eq. (ii), we get array rlrl + +2 g & =0 \ & g = p / 2 array Adding Eqs. (ii) and (iii), we get arra

Question 7

The locus of the mid-point of the chord if contact of tangents drawn from points lying on the straight line 4 x-5 y=20 to the circle x^2+y^2=9 is

Accepted Answer

Let P (t, 4 t-20 5 ) be a point on the line 4 x-5 y=20. Then the chord of contact of tangents drawn from P to the circle x^2+y^2=9 is t x+ ( 4 t-20 5 ) y=9 (i) Let (h, k) be the mid-point of this chord of contact, then its equation is also h x+k y=h^2+k^2 (ii) [ . using .T=S^ ] Clearly Eqs. (i) and (ii) represent the same line aligned & t h = 4 t-20 5 k = 9 h^2+k^2 \ & t h = 9 h^2+k^2 and t h = 4 t-20 5 k \ & t= 9 h h^2+k^2 and t= 20 h 4 h-5 k \ & 9 h h^2+k^2 = 20 h 4 h-5 k \ & h \ 20 (h^2+k

Question 8

The equation of the circle, which touches the line y=5 and passes through (-1,2) and (1,2) is

Accepted Answer

The centre of the circle is on the perpendicular bisector of the line joining (-1,2) and (1,2), which is the y-axis. The ordinate of the centre is given by array l (5-y)^ 2 =1+(y-2)^ 2 \ y= 10 3 array Hence, eq. of the circle is: array l x^ 2 + (y- 10 3 )^ 2 = ( 5 3 )^ 2 \ 9 x^ 2 +9 y^ 2 -60 y+75=0 array

Question 9

If a chord of the circle x^ 2 +y^ 2 =8 makes equal intercets of length a on the coordinate axes, then

Accepted Answer

Since the chord makes equal intercepts of length a on the coordinate axes. So, its equation can be written as x y = a . This line meets the given circle at two distinct points. So, length of the perpendicular from the centre (0,0) of the given circle must be less than the radius. i.e. | a 2 |< 8 a ^ 2 <16 | a |<4

Question 10

In the given figure, the equation of the larger circle is x^ 2 +y^ 2 +4 y-5=0 and the distance between centres is 4 . Then the equation of smaller circle is

Accepted Answer

We have x^ 2 +y^ 2 +4 y-5=0. Its centre is C _ 1 (0,-2) r _ 1 = 4+5 =3. Let C _ 2 ( ~h , k ) be the centre of the smaller circle and its radius r _ 2 . Then C _ 1 C _ 2 =4 h ^ 2 +( k +2)^ 2 =3+ r _ 2 =4 r _ 2 =1 But k = r _ 2 =1 [it touches x -axis From eq (1), 4= h ^ 2 +(1+2)^ 2 16= h ^ 2 +9 h ^ 2 =7 h = 7 Since h >0 h = 7 Hence, required circle is (x- 7 )^ 2 +(y-1)^ 2 =1

Question 11

If the coordinates at one end of a diameter of the circle x^ 2 +y^ 2 -8 x-4 y+c=0 are (-3,2) , then the coordinates at the other end are

Accepted Answer

The centre of the given circle is C (4,2) Let A (-3,2) If ( , ) are the coordinates of the other end of the diameter, then, as the middle ploint of the diameter is the centre, -3 2 =4 and +2 2 =2 =11, =2 Thus, the coordinates of the other end of diameter are (11,2)

Circle — Previous Year Questions & Solutions