A circle equation given ( x^2+y^2+4x-10y+11). Find the locus( geometrical place) of all the points which the length of the tangent between them and the circle is equal to the distance betwen them and the point (7,2)
If the given circle is $\displaystyle x^2+y^2+4x-10y+11=0$ then we may complete the square on the variables to obtain the standard form:
$\displaystyle (x+2)^2+(y-5)^2=(3\sqrt{2})^2$
Now, for some point $\displaystyle (x_0,y_0)$ outside the circle, we may form a right triangle by drawing 3 line segments:
One from the center of the circle to the tangent point $\displaystyle (x_T,y_T)$ whose length is equal to the radius of the circle, one from the tangent point to $\displaystyle (x_0,y_0)$, and one from the center of the circle to $\displaystyle (x_0,y_0)$. So, by Pythagoras, we find:
$\displaystyle (x_0-x_T)^2+(y_0-y_T)^2=(x_0+2)^2+(y_0-5)^2-18$
Now, equating this to the distance between $\displaystyle (x_0,y_0)$ and $\displaystyle (7,2)$ we find:
$\displaystyle (x_0+2)^2+(y_0-5)^2-18=(x_0-7)^2+(y_0-2)^2$
Expanding and simplifying, we will find the required locus.
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