Question

A chord \[AB\] of the circle ${x^2} + {y^2} = {a^2}$ touches the circle \[{x^2} + {y^2} - 2ax = 0\]. Find the locus of the point of intersection of tangent \[A\] and \[B\].

Answer 1

Hint: For the circle ${x^2} + {y^2} = {a^2}$, the equation of the chord of the contact is ${x_1}x + {y_1}y = {a^2}$.
We are going to use the formula to find the tangent point and substitute it in the formula for calculating the distance and with the help of this we will find the locus.
Formula for calculating the distance from a point a=$\dfrac{{a{x_1} + b{y_1} + c}}{{\sqrt {{a^2} + {b^2}} }}$

Complete step-by-step answer:
We know, the equation of the circle is ${x^2} + {y^2} + 2hx + 2fy + c = 0$
Let point \[P\left( {x,{\text{ }}y} \right)\] be the tangent such that \[T = 0\].
Now let us use the condition of tangency $x{x_1} + y{y_1} + g(x + {x_1}) + f(y + {y_1}) + c = 0$
Let us suppose that \[P\left( {h,{\text{ }}k} \right)\] is the point of intersection from the equation of the chord of the circle ${x^2} + {y^2} = {a^2}$.
Let the chord \[AB\] be the chord of the contact of the tangent from point \[P\] to the circle. So, we can assume its equation as ${x_1}x + {y_1}y = {a^2}$
Let us put the values \[\left( {h,{\text{ }}k} \right)\] in the equation ${x_1}x + {y_1}y = {a^2}$ then we will get $hx + ky = {a^2}$.
 If the line $hx + ky = {a^2}$ touches the circle \[{x^2} + {y^2} - 2ax = 0\] then, $hx + ky - {a^2} = 0$ is also the tangent of \[{x^2} + {y^2} - 2ax = 0\]
Now, let us calculate the distance from this point by using the above given formula, we get,
$a = \dfrac{{a{x_1} + b{y_1} + c}}{{\sqrt {{a^2} + {b^2}} }}$
Here $a = h,b = k$
By substituting the values in the above equation, we get the following equation:
\[a = \left| {\dfrac{{ha + k0 - {a^2}}}{{\sqrt {{h^2} + {k^2}} }}} \right|\].....(i)
Let us solve the equation so that we get the following,
\[a = \dfrac{{ah - {a^2}}}{{\sqrt {{h^2} + {k^2}} }}\]
\[a = \dfrac{{a(h - a)}}{{\sqrt {{h^2} + {k^2}} }}\]
Let us now, we will cancel out “a” on both side, we get,
$\sqrt {{h^2} + {k^2}} = \left| {h - a} \right|$
Now we square on both sides, therefore we will get
${h^2} + {k^2} = {(h - a)^2}$
We had to find the locus on \[P\left( {h,{\text{ }}k} \right)\] so we will replace \[\left( {h,{\text{ }}k} \right)\] by \[\left( {x,{\text{ }}y} \right)\] to get the locus
Therefore, the locus of point of intersection is $P(h,k) = {(x - a)^2}$

Additional Information:
The set of points that satisfy some specific conditions and properties is called locus.
The chord that joins the two points of tangents, which are drawn outside the circle to form a conic section, is called the chord of contact.
For generic circles, the equation of the chord of the contact is : $x{x_1} + y{y_1} + g(x + {x_1}) + f(y + {y_1}) + c = 0$

Note: The equation of the tangent and the equation of the chord of the contact on the point of the circle or both are \[T = 0\]. There is a difference, in tangents the (${x_1},{y_1}$) points lie on the circle, whereas in the chord of the contact these points (${x_1},{y_1}$) lie outside the circle.