Equation of radical axis of circles {x^2} + {y^2} - 3x - 4y + 5 = 0 , 2{x^2} + 2{y^2} - 10x - 12y +

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10-1.Circle and System of Circles

medium

Equation of radical axis of the circles ${x^2} + {y^2} - 3x - 4y + 5 = 0$, $2{x^2} + 2{y^2} - 10x$$ - 12y + 12 = 0$ is

A

$2x + 2y - 1 = 0$

B

$2x + 2y + 1 = 0$

C

$x + y + 7 = 0$

D

$x + y - 7 = 0$

Solution

(a) Circles are,

${S_1} = {x^2} + {y^2} – 3x – 4y + 5 = 0$ …..$(i)$

and ${S_2} = 2{x^2} + 2{y^2} – 10x – 12y + 12 = 0$

or ${S_2} = {x^2} + {y^2} – 5x – 6y + 6 = 0$ …..$(ii)$

$\therefore $ Equation of radical axis is ${S_1} – {S_2} = 0$

$ \Rightarrow $ $2x + 2y – 1 = 0$.

Standard 11

Mathematics

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