- Home
- Standard 11
- Mathematics
10-1.Circle and System of Circles
hard
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
A
${x^2} + {y^2} - 2ax = 0$
B
${x^2} + {y^2} + 2ax = 0$
C
${x^2} + {y^2} + 2ay = 0$
D
${x^2} + {y^2} - 2ay = 0$
Solution
(a) Let equation of circle be ${x^2} + {y^2} + 2gx + 2fy + c = 0$.
Since it passes through $(2a,\;0)$, so
$4{a^2} + 4ag + c = 0$.….$(i)$
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 $ag + c + a = 0$.….$(ii)$
From $(i)$ and $(ii),$ we get $g$ and $c$.
Hence equation is ${x^2} + {y^2} – 2ax = 0$.
Standard 11
Mathematics
