Question

The radius of a circle with centre \(O\) is \(\sqrt{50}\) cm. \(A\) and \(C\) are two points on the circle, and \(B\) is a point inside the circle. The length of \(AB\) is \(6\) cm, and the length of \(BC\) is \(2\) cm. The angle \(\angle ABC\) is a right angle. Find the square of the distance \(OB\).

• A. 26
• B. 25
• C. 24
• D. 23
• E. 22

Hint:

For a right triangle, the midpoint of the hypotenuse is equidistant from all three vertices (circumcenter), giving \(BM=\fracAC2\). The distance from the circle’s center to a chord \(d\) satisfies \(d=\sqrtR^2-(\tfracL2)^2\), where \(L\) is the chord length.

Answer 1

The Correct Option is A

Step 1: Since \(\angle ABC=90^\circ\), the circumcenter of \(\triangle ABC\) is the midpoint \(M\) of the hypotenuse \(AC\). Hence \[ AM=CM=\frac{AC}{2}, \qquad BM=\frac{AC}{2}. \] Given \(AB=6\), \(BC=2\) (legs), so \[ AC=\sqrt{AB^{2}+BC^{2}}=\sqrt{6^{2}+2^{2}}=\sqrt{40}=2\sqrt{10}, \] and therefore \(BM=\frac{AC}{2}=\sqrt{10}. \] Step 2: Let the radius of the given circle be \(R=\sqrt{50}\). For the chord \(AC\) of length \(\sqrt{40}\), the perpendicular distance from \(O\) to the chord is \[ OM=\sqrt{R^{2}-\left(\frac{AC}{2}\right)^{2}}=\sqrt{50-10}=\sqrt{40}=2\sqrt{10}. \] Step 3: Place \(M\) at the origin with \(AC\) on the \(x\)-axis. Then \(B\) lies on the circle \(MB=\sqrt{10}\) and \(O\) lies on the line perpendicular to \(AC\) through \(M\) with \(OM=2\sqrt{10}\). Since \(B\) is inside the big circle, \(B\) and \(O\) must be on the same side of \(AC\). Hence \[ OB^{2}=OM^{2}+MB^{2}-2\cdot OM\cdot MB\cdot \cos 0^\circ = (2\sqrt{10})^{2}+(\sqrt{10})^{2}-2(2\sqrt{10})(\sqrt{10}) =40+10-40=26. \] Thus, \(\boxed{OB^{2}=26}\).