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  • If a circle passes through the point (a, b) and cuts the circle <img alt="\mathrm{x^2+y^2=k^2}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Bx%5E2&plus;y%5E2%3Dk%5E2%7D

If a circle passes through the point (a, b) and cuts the circle \mathrm{x^2+y^2=k^2} orthogonally, then the equation of the locus of its centre is

Option: 1

\mathrm{2 a x+2 b y-\left(a^2+b^2+k^2\right)=0}


Option: 2

\mathrm{2 a x+2 b y-\left(a^2-b^2+k^2\right)=0}


Option: 3

\mathrm{x^2+y^2-3 a x-4 b y+\left(a^2+b^2-k^2\right)=0}


Option: 4

\mathrm{x^2+y^2-2 a x-3 b y+\left(a^2+b^2-k^2\right)=0}


Answers (1)

best_answer

Given circle is \mathrm{x^2+y^2=k^2}                      ...(i)

Let the equation of the circle be

\mathrm{ x^2+y^2+2 g x+2 f y+c=0 }               ...(ii)

If (i) & (ii) are orthogonal.

\mathrm{ \begin{aligned} & \Rightarrow 2 g_1 g_2+2 f_1 f_2=c_1+c_2 \\\\ & \Rightarrow 2 g(0)+2 f(0)=c-k^2 \\\\ & \Rightarrow c=k^2 \end{aligned} }                          ...(iii)

Now the circle (ii) is passing through (a, b)

\mathrm{ \begin{aligned} & \therefore \quad a^2+b^2+2 g a+2 f b+k^2=0 \\\\ & \therefore \quad 2(-g) a+2(-f) b=a^2+b^2+k^2 \end{aligned} }(Using (iii) also)

Locus of the centre \mathrm{(-g,-f)} of (ii) is

\mathrm{ 2 a x+2 b y-\left(a^2+b^2+k^2\right)=0 . }

Posted by

himanshu.meshram

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