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

The center of circle S = 0 lies on the line 2x – 2y + 9 = 0 and S = 0 cuts orthogonally the circle \mathrm{x^2+y^2=4} The circle S = 0 passes through two fixed points, find their co-ordinates.

 

Option: 1

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


Option: 2

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


Option: 3

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


Option: 4

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


Answers (1)

best_answer

Let \mathrm{S=x^2+y^2+2 g x+2 f y+c=0 \quad \Rightarrow \text { Centre }=(-g,-f) \text { lies on } 2 x-2 y+9=0 \text {. }}

\mathrm{\therefore-2 \mathrm{~g}+2 \mathrm{f}+9=0}

\mathrm{\begin{aligned} & =0 \text { cuts } x^2+y^2-4=0 \text { orthogonally. } \\ & \therefore 2 \mathrm{~g} \cdot 0+2 \mathrm{f} \cdot 0=\mathrm{c}-4 \\ & =4 . \end{aligned}}

 ∴S = 0 becomes \mathrm{x^2+y^2+(2 f+9) x+2 f y+4=0 \text { or }\left(x^2+y^2+9 x+4\right)+2 f(x+y)=0}

This represents a family of circles passing through the points of intersection of \mathrm{x^2+y^2+9 x+4=0 \text { and } x+y=0 \Rightarrow}Solving these equations we get \mathrm{(\mathrm{x}, \mathrm{y})=(-4,4),(-1 / 2,1 / 2)}

 

Posted by

Suraj Bhandari

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