If tangents are drawn to the circle x^2+y^2=1 | vedclass

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(A) Let $(h, k)$ be the point of intersection of the tangents. The chord of contact of these tangents is the common chord of the two circles.Subtracti

Vedclass · Accepted Answer

$-\frac{18}{5}$

Vedclass · Answer

(A) Let $(h, k)$ be the point of intersection of the tangents. The chord of contact of these tangents is the common chord of the two circles.Subtracting the equations of the circles $x^2+y^2=12$ and $x^2+y^2-5x+3y-2=0$ gives the equation of the common chord:$(x^2+y^2-5x+3y-2) - (x^2+y^2-12) = 0$$-5x+3y+10=0$,which simplifies to $5x-3y-10=0$.The equation of the chord of contact of the circle $x^2+y^2=12$ with respect to the point $(h, k)$ is $hx+ky=12$,or $hx+ky-12=0$.Since both equations represent the same line,their coefficients must be proportional:$\frac{h}{5} = \frac{k}{-3} = \frac{-12}{-10} = \frac{6}{5}$.Equating the ratios for $k$:$\frac{k}{-3} = \frac{6}{5} \Rightarrow k = -\frac{18}{5}$.Thus,the ordinate of the point of intersection is $-\frac{18}{5}$.

If tangents are drawn to the circle $x^2+y^2=12$ at the points of intersection with the circle $x^2+y^2-5x+3y-2=0$,then the ordinate of the point of intersection of these tangents is

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