Let O be the center of the circle Γ, and P be a point outside of circle Γ. PA is tangential to Γ at A, and PO intersects Γ at D. If PD=14 and PA=42, what is the radius of Γ.
In this case the secant - tangent theorem says
PA^2 = PD x PC , where D in on the extension of PO as it hits the circle , making CD a diameter
Let the diameter be 2x , (radius = x)
(14)(2x+14) = 42^2
28x + 196 = 1764
28x = 1568
x = 56
To find the radius of the circle Γ, we can use a property of tangents to circles.
First, let's draw a diagram to help visualize the situation. We have circle Γ with center O, point P outside the circle, and tangents PA and PD.
Since PA is tangent to the circle at A, we have a right triangle ∆POA, where ∠POA is a right angle.
Now, we can use the Pythagorean theorem to find the length of OP, which is the radius of the circle Γ.
In ∆POA, we have:
PA^2 = OA^2 + OP^2
Given that PA = 42, we can substitute this value into the equation:
42^2 = OA^2 + OP^2
We also know that PD = 14, which means the length of OD is 14. Since OD is a radius of circle Γ, it is equal to OP.
Substituting OD = OP = 14 into the equation, we have:
42^2 = OA^2 + 14^2
Simplifying the equation, we have:
1764 = OA^2 + 196
Subtracting 196 from both sides, we get:
1568 = OA^2
To find the radius of circle Γ, we need to find the value of OA. Taking the square root of both sides, we get:
sqrt(1568) = OA
Using a calculator, we find that sqrt(1568) ≈ 39.54.
Therefore, the radius of circle Γ is approximately 39.54 units.