Question

The length of the tangent drawn from any point on the circle $ {x^2} + {y^2} + 2gx + 2fy + p = 0 $ to the circle $ {x^2} + {y^2} + 2gx + 2fy + q = 0 $ is:
A. \[\sqrt {q - p} \]
B. \[\sqrt {p - q} \]
C. \[\sqrt {q + p} \]
D. \[\sqrt {2q + p} \]

Answer 1

Hint: Here, consider a random point on the circle and find the distance of that point from the second circle. As we can see from the equation of two circles that the two circles are concentric circles. Random point or general point on the circle will satisfy the equation of the circle.

Complete step-by-step answer:
Circle: Circle is a locus of a point which is moving in a two dimensional plane around a fixed point to a fixed distance. The fixed point is called centre and the fixed distance is called radius of the circle and the distance covered by the moving point in a single revolution is called circumference of the circle.
Tangent to a circle: A tangent to a circle is a line that intersects the circle at exactly one point.

Let $ \left( {{x_1},{y_1}} \right) $ be a point on the circle $ {x^2} + {y^2} + 2gx + 2fy + p = 0 $
Since, $ \left( {{x_1},{y_1}} \right) $ is a point on the given circle then it must satisfy the equation of the circle.
Thus, $ x_1^2 + y_1^2 + 2g{x_1} + 2f{y_1} + p = 0 $
 $ x_1^2 + y_1^2 + 2g{x_1} + 2f{y_1} = - p $ …(i)
Length of tangent from $ \left( {{x_1},{y_1}} \right) $ to the circle
 $ \Rightarrow {x^2} + {y^2} + 2gx + 2fy + q = 0 $ is
 $ \Rightarrow \sqrt {x_1^2 + y_1^2 + 2g{x_1} + 2f{y_1} + q} $ From equation (i), Length of tangent from $ \left( {{x_1},{y_1}} \right) $ to the circle is $ \sqrt {q - p} $ .
So, the correct answer is “Option A”.

Note: In these types of questions, for length of tangent choose a general point of the circle. As we know the tangent to a circle makes a right angle at the point of contact. Here, equations of two concentric circles are given and can easily find the length of tangent using constant given. Here, we have used the concept that any point on the circle satisfies the equation of the circle.