Equation of radical axis of circles {x^2} + {y^2} + x - y + 2 = 0 and 3{x^2} + 3{y^2} - 4x - 12 = 0,

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10-1.Circle and System of Circles

easy

The equation of radical axis of the circles ${x^2} + {y^2} + x - y + 2 = 0$ and $3{x^2} + 3{y^2} - 4x - 12 = 0,$ is

A

$2{x^2} + 2{y^2} - 5x + y - 14 = 0$

B

$7x - 3y + 18 = 0$

C

$5x - y + 14 = 0$

D

None of these

Solution

(b) Radical axis is ${S_1} – {S_2}$

${S_1} \equiv {x^2} + {y^2} + x – y + 2 = 0$

${S_2} = {x^2} + {y^2} – \frac{4}{3}x – 4 = 0$

$ \Rightarrow {S_1} – {S_2} = \frac{7}{3}x – y + 6 = 0$

or $7x – 3y + 18 = 0$.

Standard 11

Mathematics

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(IIT-1980)