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

Option: 1

\mathrm{ 2 a x+2 b y=a^2+b^2+k^2 }


Option: 2

\mathrm{ a x+b y=a^2+b^2+k^2}


Option: 3

\mathrm{ x^2+y^2+2 a x+2 b y+k^2=0}


Option: 4

\mathrm{ x^2+y^2-2 a x-2 b y+a^2+b^2-k^2=0}


Answers (1)

best_answer

Let the equation of the circle through (\mathrm{a},(\mathrm{B}) be

\mathrm{x^2+y^2+2 g x+2 f y+c=0 \quad \quad \quad \quad \dots(i)}

then \mathrm{a^2+b^2+2 g a+2 f b+c=0 \quad \quad \quad \quad \dots(ii)}

Since (i) cuts the circle \mathrm{x^2+y^2=k^2} orthogonally, we have

\mathrm{2 \mathrm{~g} \times 0+2 \mathrm{f} \times 0=\mathrm{c}-\mathrm{k}^2 \Rightarrow \mathrm{c}=\mathrm{k}^2 }

so that from (ii), we get \mathrm{a^2+b^2+2 g a+2 f b+k^2=0}, and the locus of the centre \mathrm{(-g,-f)} of (i) is \mathrm{2 a x+2 b y-\left(a^2+b^2+k^2\right)=0}

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Gaurav

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