The equation of the circle which passing thro | vedclass

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Question

(a) Let equation of circle be ${x^2} + {y^2} + 2gx + 2fy + c = 0$.

Since it passes through $(2a,\;0)$, so

$4{a^2} + 4ag + c = 0$.&hellip;.$(i)$

Vedclass · Accepted Answer

${x^2} + {y^2} - 2ax = 0$

Vedclass · Answer

(a) Let equation of circle be ${x^2} + {y^2} + 2gx + 2fy + c = 0$.

Since it passes through $(2a,\;0)$, so

$4{a^2} + 4ag + c = 0$.&hellip;.$(i)$

Also its radical axis with ${x^2} + {y^2} = {a^2}$ is

$2gx + 2fy + c + a \equiv x - \frac{a}{2} = 0$

$ \Rightarrow \frac{{2g}}{1} = \frac{{c + a}}{{ - a/2}}$ and $f = 0$

or $ag + c + a = 0$.&hellip;.$(ii)$

From $(i)$ and $(ii),$ we get $g$ and $c$.

Hence equation is ${x^2} + {y^2} - 2ax = 0$.

The equation of the circle which passing through the point $(2a,\,0)$ and whose radical axis is $x = \frac{a}{2}$ with respect to the circle ${x^2} + {y^2} = {a^2},$ will be

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