If the chord of contact of tangents from a po | vedclass

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(B) Let the equation of the circle be $x^2 + y^2 + 2gx + 2fy + c = 0$ and point $A(x_1, y_1)$.Then the equation of the chord of contact is $xx_1 + yy_

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Cut the given circle orthogonally

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(B) Let the equation of the circle be $x^2 + y^2 + 2gx + 2fy + c = 0$ and point $A(x_1, y_1)$.Then the equation of the chord of contact is $xx_1 + yy_1 + g(x + x_1) + f(y + y_1) + c = 0$,which simplifies to $(x_1 + g)x + (y_1 + f)y + (gx_1 + fy_1 + c) = 0$.Let another point $B(x_2, y_2)$ be a point through which the chord passes,so $x_1x_2 + gx_2 + y_1y_2 + fy_2 + gx_1 + fy_1 + c = 0$ ... $(i)$.The equation of the circle having $AB$ as a diameter is $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$,which simplifies to $x^2 + y^2 - (x_1 + x_2)x - (y_1 + y_2)y + x_1x_2 + y_1y_2 = 0$ ... $(ii)$.Two circles $x^2 + y^2 + 2g_1x + 2f_1y + c_1 = 0$ and $x^2 + y^2 + 2g_2x + 2f_2y + c_2 = 0$ cut orthogonally if $2g_1g_2 + 2f_1f_2 = c_1 + c_2$.For our circles,$2g_1g_2 + 2f_1f_2 = 2g(-\frac{x_1 + x_2}{2}) + 2f(-\frac{y_1 + y_2}{2}) = -gx_1 - gx_2 - fy_1 - fy_2$.From $(i)$,$-gx_1 - gx_2 - fy_1 - fy_2 = x_1x_2 + y_1y_2 + c$.Thus,$2g_1g_2 + 2f_1f_2 = c_1 + c_2$,which implies the circles cut orthogonally.Hence,option $B$ is correct.

If the chord of contact of tangents from a point $A$ to a given circle passes through $B$,then the circle with $AB$ as a diameter will . . . . . .

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