The radical centre of the circles x^2+y^2+3x+ | vedclass

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(A) Let the equations of the circles be:$S_1: x^2+y^2+3x+2y+1=0$$S_2: x^2+y^2-x+6y+5=0$$S_3: x^2+y^2+5x-8y+15=0$The radical centre is the intersection

Vedclass · Accepted Answer

$(3,2)$

Vedclass · Answer

(A) Let the equations of the circles be:$S_1: x^2+y^2+3x+2y+1=0$$S_2: x^2+y^2-x+6y+5=0$$S_3: x^2+y^2+5x-8y+15=0$The radical centre is the intersection of the radical axes of the circles taken in pairs.The radical axis of $S_1$ and $S_2$ is given by $S_1 - S_2 = 0$:$(x^2+y^2+3x+2y+1) - (x^2+y^2-x+6y+5) = 0$$4x - 4y - 4 = 0$ $\Rightarrow x - y - 1 = 0$ $\Rightarrow x = y + 1$ (Equation $1$)The radical axis of $S_2$ and $S_3$ is given by $S_2 - S_3 = 0$:$(x^2+y^2-x+6y+5) - (x^2+y^2+5x-8y+15) = 0$$-6x + 14y - 10 = 0 \Rightarrow 3x - 7y + 5 = 0$ (Equation $2$)Substituting $x = y + 1$ into Equation $2$:$3(y + 1) - 7y + 5 = 0$$3y + 3 - 7y + 5 = 0$$-4y + 8 = 0 \Rightarrow y = 2$Substituting $y = 2$ into $x = y + 1$:$x = 2 + 1 = 3$Thus,the radical centre is $(3,2)$.

The radical centre of the circles $x^2+y^2+3x+2y+1=0$,$x^2+y^2-x+6y+5=0$ and $x^2+y^2+5x-8y+15=0$ is

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