What value of x would make rq tangent to Circle P at point Q?
Solution:
We have to find the value of x that would make the RQ tangent to the circle P at point Q.
From the given figure,
If Q is tangent to the circle P, it implies that it is perpendicular to the radius at the point of tangency.
Thus, the triangle PQR is a right triangle.
By using Pythagorean theorem,
PQ2 + RQ2 = PR2
From the figure,
PQ = 9
RQ = 12
PR = (x + 9)2
(x + 9)2 + (12)2 = (x + 9)2
By using algebraic identity,
(a + b)2 = a2 + 2ab + b2
x2 + 18x + 81 = 36 + 144
x2 + 18x + 81 = 225
x2 + 18x + 81 - 225 = 0
x2 + 18x - 144 = 0
Using quadratic formula,
\(x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}\)
Here, a = 1, b = 18 and c = -144.
\(x=\frac{-18\pm \sqrt{(18)^{2}-4(1)(-144)}}{2(1)}\\x=\frac{-18\pm \sqrt{324+576}}{2}\\x=\frac{-18\pm \sqrt{900}}{2}\\x=\frac{-18\pm 30}{2}\)
\(x=\frac{-18+30}{2}\\x=\frac{12}{2}\\x=6\)
\(x=\frac{-18-30}{2}\\x=\frac{-48}{2}\\x=-24\)
Since x = -24 is negative, it can be neglected.
Therefore, the value of x is 6.
What value of x would make rq tangent to Circle P at point Q?
Summary:
The value of x would make rq tangent to Circle P at point Q is 6.
