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# Find the condition for the line px + qy + r = 0 to touch the circle ${x^2} + {y^2} = {a^2}$. And also find the coordinates of the point of contact also. Verified
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Hint: Find the equation of tangent to the circle and equate it to the equation of the line to find the point of contact. Then substitute the point in the circle equation to find the condition.

Let us find the slope of the tangent to the circle which is $\dfrac{{dy}}{{dx}}$.
Given equation of circle is as follows:
${x^2} + {y^2} = {a^2}...........(1)$
Differentiating both sides with respect to x, we get:
$2x + 2y\dfrac{{dy}}{{dx}} = 0$
Taking 2x to the other side and simplifying, we get:
$2y\dfrac{{dy}}{{dx}} = - 2x$
Now, taking 2y to the other side and finding the slope as follows:
$\dfrac{{dy}}{{dx}} = \dfrac{{ - x}}{y}$
Let the tangent intersect the circle at the point (h, k), then we have the slope of the tangent as:
$\dfrac{{dy}}{{dx}} = \dfrac{{ - h}}{k}.........(2)$
The equation of the line having slope m and passing through a point (a, b) is given by:
$y - b = m(x - a)$
Using equation (2) for the equation of tangent in the above equation, we get:
$y - k = \dfrac{{ - h}}{k}(x - h)$
Simplifying this expression by multiplying k on both sides, we get:
$ky - {k^2} = - hx + {h^2}$
Gathering constant terms on the right-hand side, we have:
$ky + hx = {h^2} + {k^2}..........(3)$
The point (h, k) lies on the circle, therefore, using equation (1), we have:
${h^2} + {k^2} = {a^2}..........(4)$
Substituting equation (4) in equation (3), we get:
$ky + hx = {a^2}$
$ky + hx - {a^2} = 0..........(5)$
The equation of the line is given by,
$px + qy + r = 0..........(6)$
From equation (5) and (6), we have:
$\dfrac{p}{h} = \dfrac{q}{k} = - \dfrac{r}{{{a^2}}}$
Hence, the value of h and k are as follows:
$h = - \dfrac{{{a^2}}}{r}p...........(7)$
$k = - \dfrac{{{a^2}}}{r}q..........(8)$
Hence, the point of intersection is $\left( { - \dfrac{{{a^2}}}{r}p, - \dfrac{{{a^2}}}{r}q} \right)$.
Now, substitute equation (7) and (8) in equation (4).
${\left( { - \dfrac{{{a^2}}}{r}p} \right)^2} + {\left( { - \dfrac{{{a^2}}}{r}q} \right)^2} = {a^2}$
$\dfrac{{{a^4}}}{{{r^2}}}{p^2} + \dfrac{{{a^4}}}{{{r^2}}}{q^2} = {a^2}$
$\dfrac{{{a^4}}}{{{r^2}}}({p^2} + {q^2}) = {a^2}$
${p^2} + {q^2} = \dfrac{{{r^2}}}{{{a^2}}}$
Hence, the required condition is ${p^2} + {q^2} = \dfrac{{{r^2}}}{{{a^2}}}$.
Note: You can directly solve the equation without finding the slope of the tangent if you know the formula for the equation of tangent to the circle at a point (h, k), that is, $hx + ky - {a^2} = 0$.