In the diagram below, PQ is tangent at P to the circle with center O, point S is inside the circle, and QS intersects the circle at R. If , QR = RS = 3, OS = 2 and PQ = 6, then find the radius of the circle.
+499 Hey there guest! First, let's start the problem with the knowledge of a (pretty common) theorem in geometry: The power of a point theorem. There are three variations to this theorem, but in this problem, we only need to look at 1 : the variation which involves the tangent of a circle and a secant(a line intersecting two points on the circle.
Firstly, notice we can extend the line QS to intersect the other side of the circle at Z. This is where power of a point comes in handy:
By power of a point:
\(QP^{2} = QR*QZ\)
Substituting known values in:
\(6^{2} = 3 * QZ\)
Dividing by 3, we get:
\(12 = QZ\)
Then RZ = QZ - 3(since that's the length of QR),
RZ = 12 - 3 = 9
Now that we have the length of a chord of Circle O(RZ is a chord), we can use power of a point once again, except another of its variations: with two intersecting chords.
A handy trick to realize is that the circle's diameter is also a chord, meaning we could find the length with power of a point theorem.
If we draw a diameter intersecting S, we could make power of a point easier to work with. By power of a point:\((R+2)*(R-2) = RS*SZ = 3*6 = 18\)
where R is the radius of the circle
This simplifies to:
\(R^{2}-4 = 18\)
\(R^{2} = 22 \)
\(R= \sqrt{22}\)
