Math Problem Statement

The two circles below are externally tangent. A common external tangent intersects line $PQ$ at $R.$ Find $QR.$

[asy] unitsize(0.3 cm);

pair A, B, C, D, E;

A = (0,0); B = (8,0); C = (20,0); D = intersectionpoint(Circle(A,5), arc((A + C)/2, abs(A - C)/2, 0, 180)); E = intersectionpoint(Circle(B,3), arc((B + C)/2, abs(B - C)/2, 0, 180));

draw(Circle(A,5)); draw(Circle(B,3)); draw(A--D); draw(B--E);

dot("$P$", A, S); dot("$Q$", B, S);

label("$8$", (B + E)/2, W, red); label("$12$", (A + D)/2, W, red); [/asy]

Solution

We are given two circles that are externally tangent. A common external tangent intersects the line PQPQ at a point RR. We are tasked with finding the length of QRQR, given some geometric information about the circles and the configuration.

Let's break down the problem step by step:

Step 1: Geometry of the Problem

  • Circle 1 is centered at PP with radius 5.
  • Circle 2 is centered at QQ with radius 3.
  • The distance between the centers PP and QQ is 8 units.
  • A common external tangent intersects line PQPQ at point RR, and we are asked to find the length of QRQR.

Step 2: Applying Geometry of Tangents

We use the formula for the distance between the points where the external tangents meet the circles, which is given by:

d=PQ2(r1r2)2d = \sqrt{PQ^2 - (r_1 - r_2)^2}

where:

  • PQPQ is the distance between the centers of the circles (8 units),
  • r1r_1 is the radius of the first circle (5 units),
  • r2r_2 is the radius of the second circle (3 units).

Step 3: Plugging Values into the Formula

Substituting the values into the formula:

d=82(53)2=644=60=215d = \sqrt{8^2 - (5 - 3)^2} = \sqrt{64 - 4} = \sqrt{60} = 2\sqrt{15}

This is the distance between the points where the external tangent touches each circle.

Step 4: Symmetry and Position of Point RR

Since RR lies on the line PQPQ, and the problem's symmetry suggests that the external tangent bisects the line between the two tangent points, we can now find QRQR as half of this distance.

Thus, QRQR is:

QR=2152=15QR = \frac{2\sqrt{15}}{2} = \sqrt{15}

Final Answer:

The length of QRQR is 15\boxed{\sqrt{15}}.

Would you like more details or have any questions on this problem?

Here are some related questions you can explore:

  1. How do we calculate the length of the internal tangent between two circles?
  2. What is the general formula for the distance between the external tangents of two circles?
  3. How does the point of tangency change if the circles are internally tangent instead of externally tangent?
  4. Can we find the equation of the common external tangent for these circles?
  5. How would the solution change if the radius of one circle were larger than the other?

Tip: When dealing with tangents, symmetry can often simplify calculations by reducing the problem to simpler geometric forms.

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Math Problem Analysis

Mathematical Concepts

Geometry
Circles
Tangents
Pythagorean Theorem

Formulas

Distance between points of tangency on external tangents: d = √(PQ^2 - (r1 - r2)^2)

Theorems

Pythagorean Theorem
Geometry of Circles
Tangent Properties

Suitable Grade Level

Grades 9-12