If the radical centre of the circles x^2+y^2- | vedclass

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Question

(B) The given circles are:
$S_1 \equiv x^2+y^2-8x-2y+8=0$
$S_2 \equiv x^2+y^2+6x+8y-24=0$
$S_3 \equiv x^2+y^2-2x+2y+2=0$
The radical axis of $S_1$

Vedclass · Accepted Answer

$10$

Vedclass · Answer

(B) The given circles are:
$S_1 \equiv x^2+y^2-8x-2y+8=0$
$S_2 \equiv x^2+y^2+6x+8y-24=0$
$S_3 \equiv x^2+y^2-2x+2y+2=0$
The radical axis of $S_1$ and $S_2$ is given by $S_1 - S_2 = 0$:
$(x^2+y^2-8x-2y+8) - (x^2+y^2+6x+8y-24) = 0$
$-14x - 10y + 32 = 0 \Rightarrow 7x + 5y = 16 \quad \dots(1)$
The radical axis of $S_2$ and $S_3$ is given by $S_2 - S_3 = 0$:
$(x^2+y^2+6x+8y-24) - (x^2+y^2-2x+2y+2) = 0$
$8x + 6y - 26 = 0 \Rightarrow 4x + 3y = 13 \quad \dots(2)$
To find the radical centre $(a, b)$,we solve the system of equations $(1)$ and $(2)$:
From $(2)$,$y = \frac{13-4x}{3}$. Substituting this into $(1)$:
$7x + 5(\frac{13-4x}{3}) = 16$
$21x + 65 - 20x = 48$
$x = -17$
Substituting $x = -17$ into the expression for $y$:
$y = \frac{13 - 4(-17)}{3} = \frac{13 + 68}{3} = \frac{81}{3} = 27$
Thus,the radical centre $(a, b) = (-17, 27)$.
Therefore,$a+b = -17 + 27 = 10$.

If the radical centre of the circles $x^2+y^2-8x-2y+8=0$,$x^2+y^2+6x+8y-24=0$,and $x^2+y^2-2x+2y+2=0$ is $(a, b)$,then $a+b=$

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