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Circle

Section: Mathematics

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Question : 32 of 89

Marks: +1, -0

From a point

P

x^2+y^2=4

, two tangents are drawn to the circle

x^2+y^2-6 x-6 y+14=0

A

B

are the points of contact of those lines, then the locus of the centre of the circle passing through the points

P, A

B

[AP EAPCET 21 May 2025 S1]

$x^2+y^2-3 x-3 y+4=0$
$2 x^2+2 y^2+6 x+6 y-7=0$
$x^2+y^2+3 x+3 y-4=0$
$2 x^2+2 y^2-6 x-6 y+7=0$

Solution:

Circle 1 given

x^2+y^2=4

Circle 2 is given,

x^2+y^2-6 x-6 y+14=0

Centre

(3,3)

radius

=2

Let

P (x_1, y_1)

be a point on circle 1 .

\because

Equation of chord of contact

A B

is

\;x x_1+y y_1-3 (x+x_1) -3 (y+y_1) +14=0

\;(x_1-3) x+ (y_1-3) y-3 x_1-3 y_1+14=0

Now, equation of circle passing through

P, A

and

B

is

x^2+y^2-6 x-6 y+14+\lambda [ (x_1-3) x+ (y_1-3) y-3 x_1-3 y_1+14] =0

We have this circle passes through

P (x_1, y_1)

\Rightarrow x_1^2+y_1^2-6 x_1-6 y_1+14+\lambda [ (x_1-3) x_1.

+ (y_1-3) y_1-3 x_1-3 y_1+14] =0

\Rightarrow x_1^2+y_1^2-6 x_1-6 y_1+14+\lambda [x_1^2+y_1^2.

-3 x_1-3 y_1-3 x_1-3 y_1+14] =x

Put

x_1^2+y_1^2=4

\;\;\;\Rightarrow 4-6 x_1-6 y_1+14

\;\;\;+\lambda (4-6 x_1-6 y_1+14) =0

\;\Rightarrow 18-6 x_1-6 y_1+14+\lambda (18-6 x_1.

\;\;\;-6 y_1) =0

\;\because 1+\lambda=0

\;\lambda=-1

\Rightarrow

Equation of circle be

\;(x^2+y^2-6 x-6 y+14)

\;- [ (x_1-3) x+ (y_1-3) y.

\;-3 x_1-3 y_1+14] =0

\;\Rightarrow x^2+y^2- (3+x_1) x- (3+y_1) y

\;+3 x_1+3 y_1=0

\;\;\text{ Centre }\;\left( \;\frac{3+x_1}{2}, \;\frac{3+y_1}{2} \right)

\;h=\;\frac{3+x_1}{2} \;\text{ and }\; k=\;\frac{3+y_1}{2}

\;x_1=2 h-3 \;\; y_1=2 k-3

\;x_1^2+y_1^2=4

\;(2 h-3)^2+(2 k-3)^2=4

\;4 h^2+4 k^2-12 h-12 k+14=0

\;h^2+k^2-3 h-3 k+\;\frac{7}{2}=0

\because

Taking locus of

(h, k)

, we get

2 x^2+2 y^2-6 x-6 y+7=0

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