The length of the tangent drawn from any poin | vedclass

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(A) The given circles are $C_1 \equiv x^2+y^2+2gx+2fy+c_1=0$ and $C_2 \equiv x^2+y^2+2gx+2fy+c_2=0$. Since both circles have the same center $(-g, -f)

Vedclass · Accepted Answer

$\sqrt{c_1-c_2}$

Vedclass · Answer

(A) The given circles are $C_1 \equiv x^2+y^2+2gx+2fy+c_1=0$ and $C_2 \equiv x^2+y^2+2gx+2fy+c_2=0$. Since both circles have the same center $(-g, -f)$,they are concentric. Let $A$ be a point on $C_1$ and $T$ be the point of tangency on $C_2$. Let $O$ be the common center. The radius of $C_1$ is $r_1 = \sqrt{g^2+f^2-c_1}$ and the radius of $C_2$ is $r_2 = \sqrt{g^2+f^2-c_2}$. In the right-angled triangle $	riangle OTA$,the length of the tangent $AT$ is given by: $AT = \sqrt{OA^2 - OT^2} = \sqrt{r_1^2 - r_2^2}$ $AT = \sqrt{(g^2+f^2-c_1) - (g^2+f^2-c_2)} = \sqrt{c_2-c_1}$. Note: For the tangent to exist,$r_1 > r_2$,which implies $c_1 < c_2$. Thus,the length is $\sqrt{c_2-c_1}$.

The length of the tangent drawn from any point on the circle $x^2+y^2+2gx+2fy+c_1=0$ to the circle $x^2+y^2+2gx+2fy+c_2=0$ is

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