Test Index

Circles Part 1

Section: Mathematics

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Question : 32 of 100

Marks: +1, -0

The circle passing through the intersection of the circles,

x^{2}+y^{2}-6x=0

x^{2}+y^{2}-4y=0

having its centre on the line,

2x-3y+12=0

also passes through the point:

[4 Sep 2020 Shift 2]

(1,-3)
(-1,3)
(-3,1)
(-3,6)

Solution:

Let

S

be the circle pasing through point of intersection of

S_{1}\ \&\ S_{2}

\therefore S = S_{1} + \lambda S_{2} = 0

\Rightarrow S:(x^{2}+ y^{2}-6x)

+ \lambda(x^{2}+ y^{2}-4y)=0

\Rightarrow S: x^{2}+ y^{2}-\frac{6}{1+\lambda} x

- \frac{4\lambda}{1+\lambda} y=0

...(1)

Centre

\left(\frac{3}{1+\lambda}, \frac{2\lambda}{1+\lambda}\right)

lies on

2x-3y+12=0 \Rightarrow \lambda = -3

put in

(1) \Rightarrow S: x^{2}+ y^{2}+3x-6y=0

Now check options point (-3,6) lies on S.

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