Question

The centre of a circle lies on the y-axis. If it passes through the points \( (-4, 3) \) and \( (3, -4) \), then its radius is:

• A. \( 7\sqrt{2} \)
• B. 4
• C. \( 4\sqrt{2} \)
• D. 5
• E. \( 5\sqrt{2} \)

Hint:

In problems involving circles, the distance from the centre to any point on the circle is always the radius. Use the distance formula to find the radius by equating the distances from the centre to two given points on the circle.

Answer 1

The Correct Option is D

Let the centre of the circle be \( C(0, r) \), where \( r \) is the radius, as the centre lies on the y-axis.
The distance between the centre \( C(0, r) \) and a point on the circle, say \( (-4, 3) \), gives the radius of the circle. Using the distance formula: \[ {Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \] For the point \( (-4, 3) \), the distance from the centre \( C(0, r) \) is: \[ {Radius} = \sqrt{(-4 - 0)^2 + (3 - r)^2} = \sqrt{16 + (3 - r)^2} \] Similarly, the distance between the centre \( C(0, r) \) and the second point \( (3, -4) \) gives the same radius: \[ {Radius} = \sqrt{(3 - 0)^2 + (-4 - r)^2} = \sqrt{9 + (-4 - r)^2} \] Now we equate the two expressions for the radius: \[ \sqrt{16 + (3 - r)^2} = \sqrt{9 + (-4 - r)^2} \] Squaring both sides: \[ 16 + (3 - r)^2 = 9 + (-4 - r)^2 \] Expanding both sides: \[ 16 + (9 - 6r + r^2) = 9 + (16 + 8r + r^2) \] Simplifying: \[ 16 + 9 - 6r + r^2 = 9 + 16 + 8r + r^2 \] \[ 25 - 6r = 25 + 8r \] Solving for \( r \): \[ -6r = 8r \] \[ r = 5 \] Thus, the radius of the circle is 5.
Thus, the correct answer is option (D), 5.