Question

Suppose two tangents \(PA\) and \(PB\) are drawn to the circle centered at \(C(1,2)\) from the point \(P(16,7)\). If the area of the quadrilateral \(PACB\) is \(75\) square units, then the radius of the circle is

• A. \(5\)
• B. \(25\)
• C. \(225\)
• D. \(\sqrt5\)

Hint:

For tangents drawn from an external point to a circle, the tangent lengths are equal and the radius is perpendicular to the tangent at the point of contact. The area of the quadrilateral formed is \[ \text{Area}=r\times \text{tangent length}. \]

Answer 1

The Correct Option is A

Step 1: Find the distance \(PC\).
The centre of the circle is \[ C(1,2) \] The external point is \[ P(16,7) \] Using distance formula, \[ PC=\sqrt{(16-1)^2+(7-2)^2} \] \[ PC=\sqrt{15^2+5^2} \] \[ PC=\sqrt{225+25} \] \[ PC=\sqrt{250}=5\sqrt{10} \]

Step 2: Relate the tangent length and radius.
Let the radius of the circle be \[ r \] Since \(PA\) and \(PB\) are tangents from the same external point, \[ PA=PB \] Also, radius is perpendicular to the tangent at the point of contact.
So, \[ CA\perp PA \] and \[ CB\perp PB \] Thus, triangles \(PAC\) and \(PBC\) are right triangles.

Step 3: Express the area of quadrilateral \(PACB\).
The quadrilateral \(PACB\) is made up of two right triangles: \[ \triangle PAC \] and \[ \triangle PBC \] Area of \(\triangle PAC\) is \[ \frac{1}{2}\cdot PA\cdot r \] Area of \(\triangle PBC\) is \[ \frac{1}{2}\cdot PB\cdot r \] Since \(PA=PB\), total area is \[ \frac{1}{2}PA\cdot r+\frac{1}{2}PA\cdot r \] \[ =PA\cdot r \] Given area is \[ 75 \] Hence, \[ PA\cdot r=75 \]

Step 4: Use the right triangle relation.
In right triangle \(PAC\), \[ PC^2=PA^2+r^2 \] Since \[ PC=5\sqrt{10}, \] we get \[ (5\sqrt{10})^2=PA^2+r^2 \] \[ 250=PA^2+r^2 \] Also, \[ PA\cdot r=75 \] Now, \[ (PA^2+r^2)^2=(PA^2-r^2)^2+4P A^2r^2 \] But a simpler way is to check the radius from the options.
If \[ r=5, \] then \[ PA=\frac{75}{5}=15 \] Now, \[ PA^2+r^2=15^2+5^2 \] \[ =225+25=250 \] This matches \[ PC^2=250 \] So, \[ r=5 \]

Step 5: Final conclusion.
Therefore, the radius of the circle is \[ \boxed{5} \]