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  • A circle passes through the origin and has its centre on If it cuts <img alt="\mathrm{x^2

A circle passes through the origin and has its centre on \mathrm{y=x. } If it cuts \mathrm{x^2+y^2-4 x-6 y+10=0} orthogonally, then the equation of the circle is
 

Option: 1

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


Option: 2

\mathrm{x^2+y^2-6 x-4 y=0}


Option: 3

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


Option: 4

\mathrm{x^2+y^2+2 x+2 y=0}


Answers (1)

best_answer

Let the required circle be \mathrm{x^2+y^2+2 g x+2 f y+c=0}          \mathrm{....(i)}

This passes through (0, 0), therefore c = 0

\mathrm{\text { The centre }(-g,-f) \text { of (i) lies on } y=x \text {, hence } g=f \text {. }}

Since (i) cuts the circle \mathrm{x^2+y^2-4 x-6 y+10=0} orthogonally, therefore

\mathrm{ 2(-2 g-3 f)=c+10 }

\mathrm{\Rightarrow-10 g=10 \Rightarrow g=f=-1 } \mathrm{\text (g=f \text { and } c=0 \text { ). } } Hence the required circle is

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

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