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Solved Subjective Questions on Circle Set 1

Posted on - 08-02-2017

JEE Math Circle

IIT JEE

Example 1.

Find the equation of the circle with centre on the line 2x + y = 0 and touching the lines 4x - 3y + 10 = 0 and 4x - 3y - 30 = 0.

Solution

Since the two lines are tangents to the given circle and they are parallel, we have

Note: We know that if ax + by + c = 0 and ax + by + c¢ = 0are two parallel tangents to a circle, then equation of the line parallel to given lines and passing through the centre is given by ax + by + .

Hence the centre of the circle lies on 4x - 3y - 10 = 0

Also the centre lies on 2x + y = 0

Hence the coordinates of the centre are (1, -2).


&⇒
The equation of the circle is (x - 1)2 + (y + 2)2 = 16.

Example 2

Find the locus of the mid-points of the chords of the circle
x2 + y2 - 2x - 6y - 10 = 0 which pass through the origin.

Solution

Let (h, k) be the coordinates of the mid point.

Equation of the chord whose mid point is (h, k) is

xh + yk - (x + h) - 3(y + k) - 10 = h2 + k2 - 2h - 6k - 10 (using T = S1)

i.e., h2 + k2 - h(x + 2) - k(6 + y) + x + h + 3y + 3k = 0.

This chord passes through the origin (0, 0). .

Hence h2 + k2 - h - 3k = 0

Hence the required locus is x2 + y2 - x - 3y = 0.

Alternative:

If m1 is the slope of the chord, then

If m2 is the slope of the perpendicular to the chord then where (1, 3) is

the centre of the given circle. But m1m2 = -1 .


&⇒
h2 + k2 - h - 3k = 0

Hence the required locus is x2 + y2 – x – 3y = 0.

Example 3.

Find the equations of the tangents from the point A(3, 2) to the circle
x2 + y2 = 4 and hence find the angle between the pair of tangents.

Solution

The equation of the pair of tangents from (3, 2) is given by T2 = SS1


&⇒
(3x + 2y - 4)2 = (x2 + y2 - 4)(9 + 4 - 4)

i.e., 9x2 + 4y2 +16+12xy -16y-24x=9x2+ 9y2–36.


&⇒
5y2 + 24x + 16y - 12xy - 52 = 0

Here a = 0, b = 5 h = -6. .

Hence angle q between the tangents is given by

or tanq =

or tanq =
&⇒
q = tan-1.

Hence the angle between the pair of tangents is

tan-1.

Example 4.

Find the locus of middle points of chords of the circle (x–2)2 + (y –3)2 = a2, which subtends a right angle at the point (b, 0).

Solution

Let M (h, k) be the middle points of the chord AB. Also ∠APB = 90° and AM = PM.

We have AM =

=

PM =


&⇒
b2– 2bh+h2 + k2 = a2 – (h–2)2 – (k – 3)2


&⇒
h2+k2–h(b–2)– 3k +

Hence locus of (h, k) is

x2 + y2 – x (b – 2) – 3y +

which is a circle.

Example 5.

If a straight line through C(-Ö8, Ö8), making an angle of 135° with the x-axis, cuts the circle x = 5 cos q, y = 5 sin q, in points A and B, find the length of the segment AB.

Solution

The given circle is x2 + y2 = 25 . . .(1).

Equation of line through C is


&⇒

Any point on this line is

If the point lies on the circle x2 + y2 = 25, the


&⇒


&⇒

&⇒
r2 + 8r + 16 – 25 = 0


&⇒
r2 – 9r + r – 9 = 0
&⇒
r = 9, ­–1

AB = 9 – (–1) = 10.

 
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