Question

If 5x-12y +10 = 0 and 12y – 5x+16 = 0 are two tangents of a circle, then the radius of the circle is
[a] 1
[b] 2
[c] 4
[d] 6

Answer 1

Hint: Use the fact that if two tangents to a circle are parallel to each other, then the chord formed by their points of contact is the diameter of the circle. Hence prove that the distance between the tangents is equal to the diameter of the circle. Use the fact that two lines are parallel to each other if their slopes are equal. Use the fact that the slope of the line ax+by +c= 0 is given by$\dfrac{-a}{b}$. Hence prove that the lines 5x-12y+10=0 and 12y-5x+16 =0 are parallel to each other. Use the fact that the distance between the parallel lines $ax+by+{{c}_{1}}=0$ and $ax+by+{{c}_{2}}=0$ is given by $d=\dfrac{\left| {{c}_{2}}-{{c}_{1}} \right|}{\sqrt{{{a}^{2}}+{{b}^{2}}}}$. Hence find the diameter of the circle and hence the radius of the circle.

Complete step-by-step answer:

We know that the slope of the line ax+by +c= 0 is given by$\dfrac{-a}{b}$
Hence the slope of the tangent 5x-12y+ 10 = 0 is $\dfrac{-5}{-12}=\dfrac{5}{12}$ and the slope of the tangent 12y -5x+16 = 0 is $\dfrac{-\left( -5 \right)}{12}=\dfrac{5}{12}$
We know that if the slopes of two lines are equal, then the lines are parallel.
Hence, we have the lines 5x-12y +10 =0 and 12y-5x=16 =0 are parallel to each other.
We know that that the distance between the parallel lines $ax+by+{{c}_{1}}=0$ and $ax+by+{{c}_{2}}=0$ is given by $d=\dfrac{\left| {{c}_{2}}-{{c}_{1}} \right|}{\sqrt{{{a}^{2}}+{{b}^{2}}}}$.
12y – 5x+16 = 0
Multiplying both sides by -1, we get
$5x-12y-16=0$
Hence, the equation of the two tangents are
 $\begin{align}
  & 5x-12y+10=0 \\
 & 5x-12y-16=0 \\
\end{align}$
Hence the distance between the tangents $=\dfrac{\left| 10-\left( -16 \right) \right|}{\sqrt{{{12}^{2}}+{{5}^{2}}}}=\dfrac{26}{13}=2$
We know that the distance between two parallel tangents of a circle is equal to the diameter of the circle.
Hence, we have the diameter of the circle = 2 units
Hence the radius of the circle = 1 unit.
Hence option [a] is correct.

Note: Consider a circle as shown. Lines l and m are two tangents to the circle. Join OA and OB.

As the tangent and the radius at the point of contact are parallel, we have OA is perpendicular to l and OB is perpendicular to m. Now, since l is parallel to m, we have a line perpendicular to l that is also perpendicular to m. Hence line OA if extended will be perpendicular from. Since perpendicular from the centre of a circle to a tangent of the circle passes through the point of contact, we have OA passes through B. Hence O, A and B are collinear. Hence, we have AB = 2OA.
Also, since AB is perpendicular to both l and m, we have, AB is the distance between the parallel lines l and m.
Hence the distance between the parallel tangents is equal to the diameter of the circle.