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Line segment \( QP \) is tangent to the circle.

A circle is shown with secant \( MP \) and tangent \( QP \) intersecting at point \( P \) outside of the circle. Secant \( MP \) intersects the circle at point \( N \). The length of \( QP \) is \( n \), the length of \( NP \) is 11.5, and the length of \( MN \) is 24.

What is the length of line segment \( QP \)? Round to the nearest unit.

A. 13 units
B. 17 units
C. 18 units
D. 20 units

Answer :

Answer:

QP ≈ 20 units

Step-by-step explanation:

We can find the length of line segment QP by this way:

Let:

  • O is the center of the circle
  • r is the radius of the circle

Given:

  • NP = 11.5 units
  • MN = 24 units

First, let's look at the ΔMON. Since the length of MO equals to NO, then ΔMON is an isosceles triangle, thus:

[tex]\begin{aligned}MO'&=NO'\\\\NO'&=\frac{1}{2}\times MN\\\\&=\frac{1}{2}\times24\\\\&=12\ units \end{aligned}[/tex]

Now, we look at the ΔOO'N. Since it is a right triangle and according to the Pythagorean theorem:

[tex]\begin{aligned}NO^2&=OO'^2+NO'^2\\\\r^2&=OO'^2+12^2\\\\OO'^2&=r^2-144\end{aligned}[/tex]

Next, we look at the ΔOO'P. Since it is a right triangle too and according to the Pythagorean theorem:

[tex]\begin{aligned}OP^2&=OO'^2+OP^2\\\\&=(r^2-144)+(NO'+NP)^2\\\\&=r^2-144+(12+11.5)^2\\\\&=r^2+408.25\end{aligned}[/tex]

Finally, we look at the ΔOPQ. Since QP is tangent to the circle, then QP is perpendicular to the radius, thus ΔOPQ is a right triangle with the right angle at ∠Q. According to the Pythagorean theorem:

[tex]\begin{aligned}OP^2&=OQ^2+QP^2\\\\r^2+408.25&=r^2+QP^2\\\\QP^2&=408.25\\\\QP&\approx20\ units\end{aligned}[/tex]

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