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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?

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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,

enter image description here 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.

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