P is a point outside of circle Γ. The tangent from P to Γ touches at A. A line from P intersects Γ at B and C such that m∠ACP = 120∘. If AC=16 and AP=19, find the radius of the circle.
Let O = centre of circle Γ, then
∠PAO=90°
Let O = centre of circle Γ, then
∠PAO=90°
Consider ΔPAC,
use sine rule to find ∠PAC, which
equals 13° (approx.).
∠CAO is therefore 90-∠PAC and equals 77° approx.
Since Δ CAO is isosceles, with congruent legs equal to the radius r of the circle Γ, and base length = 16, r can be solved.
To find the radius of the circle, we can first find the length of AB and BC using the given information.
Since PA is tangent to the circle, we have ∠PAB = 90°. In the right triangle PAB, we can use the Pythagorean theorem to find AB:
AB^2 = AP^2 - PA^2
AB^2 = 19^2 - (r)^2
AB^2 = 361 - r^2
Next, we need to find the length of BC. To do this, we can use the inscribed angle theorem, which states that the measure of an inscribed angle is half the measure of its intercepted arc. In this case, ∠ACP is an inscribed angle that intercepts arc BC. Since ∠ACP = 120°, we know that arc BC has a measure of 240°.
Using the arc measure, we can set up the following equation based on the circumference of the circle:
BC/∠BCA = (circumference of the circle)/(360°) = r/(2πr) = 240°/360°
Simplifying this equation, we get:
BC/r = 2/3
BC = (2/3)r
Now, we have two equations:
AB^2 = 361 - r^2
BC = (2/3)r
Since AB + BC = AC, we can substitute the values of AB and BC into this equation:
(361 - r^2) + (2/3)r = 16
Multiplying through by 3 to eliminate the fraction:
3(361 - r^2) + 2r = 48
Simplifying:
1083 - 3r^2 + 2r = 48
Rearranging and setting it equal to zero:
3r^2 - 2r + 1035 = 0
This is a quadratic equation that can be solved using the quadratic formula:
r = (-b ± √(b^2 - 4ac)) / (2a)
Using a = 3, b = -2, and c = 1035, we can substitute these values into the quadratic formula:
r = (-(-2) ± √((-2)^2 - 4(3)(1035))) / (2(3))
r = (2 ± √(4 - 12420)) / 6
r = (2 ± √(-12416)) / 6
Since the discriminant (b^2 - 4ac) is negative, the square root of a negative number is not a real number. Therefore, there is no real solution for the radius of the circle in this case.