Show that length of tangent from any point on the circle$ x^2 + y^2 + 2gx + 2fy +c=0 $ to the circle $x^2 + y^2 + 2gx + 2fy +c_1 =0 $ is $\sqrt {c}-c_1$
[Math] Length of tangent from one circle to another
circles
Related Solutions
Dividing the equation throughout by 3 is NOT a foolish attempt (a correct one instead).
The calculated answer is 12 which matches your finding.
Going through the trouble of sketching the figure in Geogebra confirms that the tangent length is 12 also. See below.
The equation of the line perpendicular to the given tangent at $B (5,-5)$ can be found to be $4x + 3y = 5$. The centre of the circle must lie on this line. The centre must also lie on the perpendicular bisector of the line joining $A$ and $B$. The equation of this perpendicular can be found by using the fact that it passes through the mid point of $AB$ and perpendicular to $AB$. The equation of this line will be $x - y = 3$. Solving the linear equation gives the centre as $(2,-1)$. The radius is $\sqrt{4^2+(-3)^2} = 5$.
So the equation of circle is $(x-2)^2 +(y+1)^2 = 25$ which can be expressed in standard form as $x^2 + y^2 - 4x + 2y -20 = 0$. This is the answer.
Best Answer
Let $C_1:x^2+y^2+2gx+2fy+c=0$ and $C_2=x^2+y^2+2gx+2fy+c_1=0$. Now note that $C_1=(x+g)^2+(y+f)^2=f^2+g^2-c$ and $C_2=(x+g)^2+(y+f)^2=f^2+g^2-c_1$. Now assuming that $c\ge c_1$, $C_1$ is contained in $C_2$ and the length of the tangent from any point on $C_1$ to $C_2$ is $r_2^2-r_1^2=\sqrt{c-c_1}$, using the fact that tangent of the circle is perpendicular to the radius at the point of tangency.