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  • The center of the circle lie on the line <img alt="\mathrm{2x-2y+9=0}" src="https:

The center of the circle \mathrm{S = 0} lie on the line \mathrm{2x-2y+9=0}  and  \mathrm{s=0} cuts orthogonally the circle \mathrm{x^{2}+y^{2}=4} . The circle \mathrm{s=0}  passes through two fixed points. Find mid-point of their co-ordinates.

Option: 1

\left ( \frac{-7}{4},\frac{7}{4} \right )


Option: 2

\left ( 3,-3 \right )


Option: 3

\left ( -2,2 \right )


Option: 4

\left ( -\frac{9}{4},\frac{9}{4} \right )


Answers (1)

best_answer

Let circle be

\mathrm{ S \equiv x^2+y^2+2 g x+2 f y+c=0}           ....(1)

Since center of this circle \mathrm{(-g,-f) lie\: on \: 2 x-2 y+9=0}

\therefore \quad-2 \mathrm{~g}+2 \mathrm{f}+9=0                            .......(2)

and the circle \mathrm{S=0} and \mathrm{x^2+y^2-4=0} cuts orthogonally.

\begin{array}{ll} \therefore & 2 \mathrm{~g} \times 0+2 \mathrm{f} \times 0=\mathrm{c}-4 \\ \therefore & \mathrm{c}=4 \end{array}             ..........(3)

Substituting the values of \mathrm{g} and \mathrm{c} from (2) \& (3) in (1) then \mathrm{x^2+y^2+(2 f+9) x+2 f y+4=0 }

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

Hence the circle \mathrm{ S=0} passes through fixed point  \left(\because \text { form } S^{\prime}+\lambda \cdot P=0\right)

\mathrm{\therefore \quad x^2+y^2+9 x+4=0\: and \: x+y=0}

After solving we get \mathrm{(-4,4) or \left(-\frac{1}{2}, \frac{1}{2}\right).}

 

 

Posted by

SANGALDEEP SINGH

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