left(0, frac{3}{4}right) is the radical centr | vedclass

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(A) The radical centre is the point of intersection of the radical axes. Since $\left(0, \frac{3}{4}\right)$ is the radical centre,it must satisfy the

Vedclass · Accepted Answer

$\left(\frac{-11}{12}, 1\right)$

Vedclass · Answer

(A) The radical centre is the point of intersection of the radical axes. Since $\left(0, \frac{3}{4}\right)$ is the radical centre,it must satisfy the equations of the radical axes $S_1-S_2=0$ and $S_2-S_3=0$. First,consider the radical axis $S_1-S_2=0$: $(x^2+y^2-2x+6y) - (x^2+y^2+2gx-2y+6) = 0$ $(-2-2g)x + 8y - 6 = 0$. Substituting $\left(0, \frac{3}{4}\right)$ into this equation: $(-2-2g)(0) + 8\left(\frac{3}{4}\right) - 6 = 0 \Rightarrow 6 - 6 = 0$. This is consistent. Next,consider the radical axis $S_2-S_3=0$: $(x^2+y^2+2gx-2y+6) - (x^2+y^2-12x+2fy+3) = 0$ $(2g+12)x + (-2-2f)y + 3 = 0$. Substituting $\left(0, \frac{3}{4}\right)$ into this equation: $(2g+12)(0) + (-2-2f)\left(\frac{3}{4}\right) + 3 = 0$ $(-2-2f)\left(\frac{3}{4}\right) = -3 $ $\Rightarrow -2-2f = -4$ $\Rightarrow 2f = 2$ $\Rightarrow f = 1$. Since $S_2$ and $S_3$ intersect orthogonally,the condition $2g_1g_2 + 2f_1f_2 = c_1 + c_2$ applies: $2(g)(-6) + 2(-1)(f) = 6 + 3$ $-12g - 2f = 9$. Substituting $f=1$: $-12g - 2(1) = 9$ $\Rightarrow -12g = 11$ $\Rightarrow g = \frac{-11}{12}$. Thus,$(g, f) = \left(\frac{-11}{12}, 1\right)$.

$\left(0, \frac{3}{4}\right)$ is the radical centre of the circles $S_1: x^2+y^2-2x+6y=0$,$S_2: x^2+y^2+2gx-2y+6=0$,and $S_3: x^2+y^2-12x+2fy+3=0$. If $S_2$ and $S_3$ intersect orthogonally,then $(g, f) =$

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