Question

If $2x-3y+5=0$ and $4x-5y+7=0$ are the equations of the normals drawn to a circle and (2,5) is a point on the given circle, then the radius of the circle is

• A. 1
• B. 2
• C. 3
• D. 4

Hint:

A key property of circles is that any line normal to the circle must pass through the center. Therefore, the intersection of any two distinct normals gives the coordinates of the center.

Answer 1

The Correct Option is B

The normals to a circle always pass through its centre.
Therefore, the centre of the circle is the point of intersection of the two given normal lines.
Line 1: $2x-3y = -5$.
Line 2: $4x-5y = -7$.
We solve this system of linear equations to find the centre $(h,k)$.
Multiply Line 1 by 2: $4x-6y = -10$.
Subtract this new equation from Line 2:
$(4x-5y) - (4x-6y) = -7 - (-10)$.
$y = 3$.
Substitute $y=3$ into Line 1:
$2x - 3(3) = -5 \implies 2x - 9 = -5 \implies 2x = 4 \implies x = 2$.
So, the centre of the circle is $C(2, 3)$.
We are given that the point $P(2, 5)$ lies on the circle.
The radius of the circle is the distance between the centre C and any point P on the circle.
Radius $r = CP$.
Using the distance formula:
$r = \sqrt{(2-2)^2 + (5-3)^2} = \sqrt{0^2 + 2^2} = \sqrt{4} = 2$.
The radius of the circle is 2.