# If tangents are drawn from two points which are equidistant from given point, then find the locus

Tangents are drawn to the circle $x^2+y^2=a^2$ from two points on the $X$ axis equidistant from the point $(k,0)$ prove locus of their intersection is $ky​^2=a^2(k-x)$.

If I take points as $(k+\alpha,0)$ and $k-\alpha,0)$ then while writing the equation of tangent, two new variable in form of slope of tangent will be introduced. So how will we find locus. Could someone suggest some approach?

For my convenience, I have changed the notation a little bit --- $a = r$ and $\alpha = R$.

Fact:- The equation of the tangent pair from P(X, Y) to the circle $x^2 + y^2 = r^2$ is $$(X^2 + Y^2 – r^2)(x^2 + y^2 – r^2) = (xX + yY – r^2)^2$$

As seen from the figure,

we have the following two equations $$(1) … ((k + R)^2 – r^2)(x^2 + y^2 – r^2) = x(k + R) – r^2)^2$$

$$(2) … ((k – R)^2 – r^2)(x^2 + y^2 – r^2) = x(k – R) – r^2)^2$$

To find the locus of the points of intersection of these tangents, we do (1) – (2).

Then, $(x^2 + y^2 – r^2)[((k + R)^2 – r^2) –((k – R)^2 – r^2)] = [x(k + R) – r^2)^2] – [x(k – R) – r^2)^2]$

Result follows after simplifying the above.