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  • If the centre of a circle S = 0 lies on the line 2x – 2y + 9 = 0 and circle S = 0 cuts <img alt="\mathrm{x^{2}+y^{2}=4}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B

If the centre of a circle S = 0 lies on the line 2x – 2y + 9 = 0 and circle S = 0 cuts \mathrm{x^{2}+y^{2}=4} orthogonally, then S = 0 passes through fixed points 

Option: 1

\mathrm{(-1,-1),(4,4)}


Option: 2

\mathrm{\left(-\frac{1}{2},+\frac{1}{2}\right),(-4,4)}


Option: 3

\mathrm{\left(-\frac{1}{2},-\frac{1}{2}\right),(4,4)}


Option: 4

none of these


Answers (1)

best_answer

Let the circle be \mathrm{x^2+y^2+2 g x+2 f y+c=0}centre is (–g, –f) which lies on the line.

\therefore The condition is\mathrm{-2 g+2 f+9=0 \Rightarrow \quad 2 g-2 f-9=0} .........(1)

The orthogonality condition gives \mathrm{0+0=c-4 \Rightarrow c=4}.On eliminating g and c,

the equation of circle gives

\mathrm{x^2+y^2+(2 f+9) x+2 f y+4=0}; 

Or \mathrm{x^2+y^2+9 x+4+f(2 x+2 y)=0}; 0; Fixed points are given by

\mathrm{ x+y=0 \: \& \: x^2+y^2+9 x+4=0 \Rightarrow x^2+x^2+9 x+4=0}

\mathrm{ \: Or \: 2 x^2+9 x-4=0 ; x=\frac{-9 \pm \sqrt{81-32}}{4}=\frac{-9 \pm \sqrt{49}}{4}}

\mathrm{=\frac{-9 \pm 7}{4}=-4,-\frac{1}{2}}

\mathrm{\therefore }The fixed points are \mathrm{\left(-\frac{1}{2}, \frac{1}{2}\right),(-4,4)}

Posted by

Divya Prakash Singh

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