Equation of a circle passing through origin and co-axial to circles {x^2} + {y^2} = {a^2} and {x^2}

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10-1.Circle and System of Circles

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The equation of a circle passing through origin and co-axial to circles ${x^2} + {y^2} = {a^2}$ and ${x^2} + {y^2} + 2ax = 2{a^2},$ is

A

${x^2} + {y^2} = 1$

B

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

C

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

D

${x^2} + {y^2} = 2{a^2}$

Solution

(c) Equation of the circle which passes through origin is ${x^2} + {y^2} + 2gx + 2fy = 0$.

Radical axis with both circles is

$2gx + 2fy + {a^2} = 0$….$(i)$

$2(g – a)x + 2fy + 2{a^2} = 0$ ….$(ii)$

Also radical axis of the two circles is

$x = \frac{a}{2} \Rightarrow f = 0$

From $(i)$ and $(ii),$ we get $\frac{{2g}}{{2(g – a)}} = \frac{1}{2}$

$\Rightarrow g = – a$

Hence circle is ${x^2} + {y^2} – 2ax = 0$.

Standard 11

Mathematics

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