Two circles of equal radius a cut orthogonall | vedclass

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(C) Let $S_1$ be the circle with centre $(2, 3)$ and radius $a$. The equation is $(x-2)^2 + (y-3)^2 = a^2$,which simplifies to $x^2 + y^2 - 4x - 6y +

Vedclass · Accepted Answer

$\left(a, \frac{5a}{3}\right)$

Vedclass · Answer

(C) Let $S_1$ be the circle with centre $(2, 3)$ and radius $a$. The equation is $(x-2)^2 + (y-3)^2 = a^2$,which simplifies to $x^2 + y^2 - 4x - 6y + 13 - a^2 = 0$.Let $S_2$ be the circle with centre $(5, 6)$ and radius $a$. The equation is $(x-5)^2 + (y-6)^2 = a^2$,which simplifies to $x^2 + y^2 - 10x - 12y + 61 - a^2 = 0$.The radical axis is given by $S_1 - S_2 = 0$:$(x^2 + y^2 - 4x - 6y + 13 - a^2) - (x^2 + y^2 - 10x - 12y + 61 - a^2) = 0$$6x + 6y - 48 = 0 \Rightarrow x + y = 8$.Since the circles cut orthogonally,$2g_1g_2 + 2f_1f_2 = c_1 + c_2$.Here $g_1 = -2, f_1 = -3, c_1 = 13 - a^2$ and $g_2 = -5, f_2 = -6, c_2 = 61 - a^2$.$2(-2)(-5) + 2(-3)(-6) = (13 - a^2) + (61 - a^2)$$20 + 36 = 74 - 2a^2$ $\Rightarrow 56 = 74 - 2a^2$ $\Rightarrow 2a^2 = 18$ $\Rightarrow a^2 = 9$ $\Rightarrow a = 3$.Substituting $a = 3$ into the options:$(A)$ $(9, 15) \Rightarrow 9 + 15 = 24 
eq 8$.$(B)$ $(6, 3) \Rightarrow 6 + 3 = 9 
eq 8$.$(C)$ $(3, 5) \Rightarrow 3 + 5 = 8$. This satisfies the equation.$(D)$ $(3, 3) \Rightarrow 3 + 3 = 6 
eq 8$.Thus,option $(C)$ is correct.

Two circles of equal radius $a$ cut orthogonally. If their centres are $(2, 3)$ and $(5, 6)$,then the radical axis of these circles passes through the point:

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