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  • A circle C passes through (2a,0) and has the line 2x-a=0 as the radical axis with the circle <img alt="\mathrm{x^2+y^2=a^2}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Bx%

A circle C passes through (2a,0) and has the line 2x-a=0 as the radical axis with the circle \mathrm{x^2+y^2=a^2}, then equation of circle is

Option: 1

x^2+y^2-a x=0


Option: 2

x^2+y^2-2 a x=0


Option: 3

x^2+y^2+2 a x=0


Option: 4

x^2+y^2+a x=0


Answers (1)

best_answer

Equation of the circle be \mathrm{x^2+y^2+2 g x+2 f y+c=0} ,it passes through (2a, 0)

\mathrm{\Rightarrow 4 a^2+4 a g+c=0 .} ------------(i)

Also, its radical axis with \mathrm{x^2+y^2=a^2 \text { is } 2 g x+2 f y+c+a^2=0}

But the radical axis is x = a/2

\mathrm{\begin{aligned} & \therefore \frac{2 g}{1}=\frac{c+a^2}{-a / 2} \text { and } f=0 \\ & \Rightarrow a g+c+a^2=0 \end{aligned}}

On solving (i) and (ii), we get

\mathrm{\begin{aligned} c & =0, g=-a \\ \text { also } f & =0 \end{aligned}}

Hence equation of circle is \mathrm{x^2+y^2-2 a x=0}

Posted by

shivangi.bhatnagar

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