Question

If length of the tangent is \(8\text{ cm}\) and the distance between the center of the circle and the external point is \(11\text{ cm}\), then the area of the circle is

• A. \(100\text{ cm}^2\)
• B. \(197.14\text{ cm}^2\)
• C. \(179.14\text{ cm}^2\)
• D. \(110.14\text{ cm}^2\)

Hint:

Radius is always perpendicular to the tangent at the point of contact. So, use \(OP^2=OT^2+PT^2\) to find the radius.

Answer 1

The Correct Option is C

Let \(O\) be the center of the circle and \(P\) be the external point. The tangent from \(P\) touches the circle at \(T\). We are given: \[ PT=8\text{ cm} \] and \[ OP=11\text{ cm}. \] Since radius is perpendicular to the tangent at the point of contact, \[ OT\perp PT. \] Therefore, triangle \(OPT\) is a right-angled triangle. Let the radius of the circle be \(r\). Then, \[ OT=r. \] Using Pythagoras theorem in triangle \(OPT\): \[ OP^2=OT^2+PT^2. \] Substitute the given values: \[ 11^2=r^2+8^2. \] \[ 121=r^2+64. \] \[ r^2=121-64. \] \[ r^2=57. \] Now area of the circle is \[ A=\pi r^2. \] So, \[ A=\pi(57). \] Taking \[ \pi=\frac{22}{7}, \] we get \[ A=\frac{22}{7}\times 57. \] \[ A=179.14\text{ cm}^2. \] Hence, the area of the circle is \[ 179.14\text{ cm}^2. \]