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Solved Subjective Questions on Circle Set 5

Posted on - 18-02-2017

JEE Math Circle

IIT JEE

Example 1.

Find the locus of the point of intersection of tangents to the circle x = acos q,
y = asinq at the point whose parametric angles differ by (i) p/3 (ii) p/2.

Solution

Let A and B be two points on circle whose parametric angle differs by q

then PA =

tan . . . . . (1)

(i) when q = p/3 then

eq. (1) becomes h2 +k2 –a2 = a2/3.

locus of point of intersection P is x2+y2 =

(ii) when q = p/2 eq.(1) becomes .

h2 +k2 –a2 = a2

locus of point of intersection P is x2 +y2 = 2a2

Example 2.

A point moves so that the sum of the squares of its distances from n fixed point is given. Prove that its locus is a circle.

Solution

Let (h, k) be the moving point and (ai, bi); i = 1, 2, 3, …, n be the fixed points. Then given (constant)


&⇒


&⇒
n ( h2 + k2) – 2h Sai – 2kSbi + S(ai2 + bi2) - l = 0

\ locus of (h, k) is x2 + y2 - 2a x - 2b y + g = 0 which is a circle

where a = , b = , g =

Example 3.

Lines 5x + 12y - 10 = 0 and 5x - 12y - 40 = 0 touch circle C1 of diameter 6. If centre of C1 lies in the first quadrant, find the equation of circle C2, which is concentric with circle C1 and cuts an intercept of length 8 on these lines.

Solution

Since the lines 5x +12y – 10 = 0 and 5x–12y– 40 =0 touch the circle, its centre lies on one of the angle bisectors of the given lines.

i.e

5x + 12y – 10 = ±(5x – 12y – 40)


&⇒
24y = –30 or 10x = 50
&⇒
or x = 5

But the centre lies in first quadrant
&⇒
x = 5

Since C­1 touches 5x + 12y – 10 = 0,


&⇒
25 + 12h – 10 = ± 39
&⇒
12h = ± 39 – 15


&⇒
12h = 24
&⇒
h = 2 also h = (not possible)

centre of C1 = (5, 2)

Let r be radius of C2
&⇒
r2 = OP2 + AP2 = 32 + 42 = 52 or r = 5

Equation of C2 is (x – 5)2 + (y – 2)2 = 52


&⇒
x2 + y2 – 10x – 4y + 4 = 0

Example 4.

A and B are two points on the x-axis and the y-axis respectively. Two circles are drawn passing through the origin and having centres at A and B respectively. Show that AB bisects the common chord of these circles.

Solution

Let A(a, 0) and B(0, b). Then the two circles are .

x2 + y2 – 2ax = 0 …(1)

and x2 + y2 – 2by = 0

Common chord is ax – by = 0 ….(2) .

Equation of circle through (1) and (2) is x2 + y2 – 2ax + l(ax – by) = 0

i.e. x2 + y2 + ax(l - 2) - l by = 0 .

This will be circle with common chord (2) as diameter if centre lies on (2) i.e.
&⇒
l =

Mid-point of common chord is

i.e.

This lie on line AB ( equation ) if

which is true

so AB passes through mid-point of common chord.

Example 5.

Prove that the locus of the centre of a circle which touches the circle
x2 + y2 – 6x – 6y + 14 = 0 externally and touches the y-axis is given by
y2 – 10x –6y + 14 = 0
. .

Solution

Centre of given circle is (3, 3) and radius = 2. .

Let r = radius and (h, k) be the centre of other circle. .

r = length of perpendicular from (h, k) on x = 0 (y-axis)

\ r = h

Equation is x2 + y2 – 2hx – 2ky – h = 0

Again r + 2 = distance between the centres


&⇒
h + 2 =


&⇒
(h + 2)2 = (h – 3)2 + (k - 3)2


&⇒
k2 – 6k – 10h + 14 = 0

\ locus of centre (h, k) is y2 - 6k – 10x + 14 = 0.

 
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