Question

$x_{2}+y_{2}+13x−3y=0$ and $2x_{2}+2y_{2}+4x−7y−25=0$ is

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Solution

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Correct option is A)

$(x_{2}+y_{2}+13x−3y)+k(2x_{2}+2y_{2}+4x−7y−25)=0$

It is given that equation of circles passes through the point $(1,1)$

So,

$x=1$ and $y=1$

Substitute these value in an above equation, we get

$(1+1+13−3)+k(2+2+4−7−25)=0$

$12−24k=0$

$k=21 $

Now, Subsitute the value of k in an first equation, we get

$(x_{2}+y_{2}+13x−3y)+21 (2x_{2}+2y_{2}+4x−7y−25)=0$

$2x_{2}+2y_{2}+15x−213y −225 =0$

$4x_{2}+4y_{2}+30x−13y−25=0$

Video Explanation

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