Basics of Circles 9/20/15. Important theorems:

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1 Basics of ircles 9/20/15 B H P E F G ID Important theorems: 1. A radius is perpendicular to a tangent at the point of tangency. PB B 2. The measure of a central angle is equal to the measure of its intercepted arc. 3. The measure of an inscribed angle is equal to ½ the measure of its intercepted arc. 4. orollary: An angle inscribed in a semicircle is a right angle. (<HGD is a right angle.) 5. Power of a Point: Given a point P (located anywhere on the plane) and a line through P which intersects some circle in two points A and B, the product (PA)(PB) is the same for any choice of the line. The value of the product is called the power of the point.

2 Now for some fun!! 1. A circle passes through (2, 2) and (9, 9) and is tangent to the positive x-axis. ompute the x-coordinate of the point of tangency. 2. A circle P, radius 1, is inscribed in a 60 sector. Find the radius of the larger circle. A B 3. Given that the radius of the larger circle is a and the radius of the smaller circle is b, and the two circles are tangent to one another and share the common external tangent, find D. D 4. Two circles with radii 1 and 4 are externally tangent. Find the sine of the angle formed by the two common external tangents. 5. In circle O, E = ED, and <AOB is a right angle. [ED] is the area of triangle ED. Find the ratio [ED] : [AOB].

3 E O D A B 6. Polygon IJKLMN is circumscribed about circle A. Measurements are given in the figure. Find the area of IJKLMN. 7. A snowman (albeit a mis-proportioned snowman) is made of three increasingly smaller circles, each inscribed in a triangle with sides of 10, 10, and 12. Find the radius of the head part of the snowman.

4 8. AB is tangent to circle O and equal to 6. B = D = 3 and OD = 2. Find the radius of circle O. 9. Square ABD contains two congruent, tangent semicircles. D = 4. A small circle is drawn tangent to the square and to both circles. Find the radius of the small circle. 10. ircles Q and R intersect in two points as shown. Find the locus of all points, P, for which the power of point P to circle Q is equal to the power of point P to circle R.

5 11. ircles P and Q intersect in R and W. W, P, R and T are collinear. TV and TS are tangent to the circles. PS = 6 and TV = 8. Find TR. 12. ircles M and N intersect with EP containing the common chord. Points A and B are the intersections of the circles with MN. If the radius of circle N is 5, the radius of circle M is 3, and PB = 1, find the value of AP.

6 Hints and Answers 1. Use the distance formula or right triangles to get the length of the secant segment and the external part of the secant segment. Use the Power of a Point Theorem. The answer is Draw the radius of the larger circle that goes through P. Now drop a radius to the point of tangency of circle P to the radius of the larger circle. Use a triangle to get the distance between the centers and then add one for the additional length. Answer: Draw the segment between the centers and form a right triangle. Answer: 2 ab 4. Draw the segment from the exterior point to the center of the larger circle. Then draw radii to the tangent. Use similar right triangles and SOH-AH-TOA. For the sine of the full angle, use the double angle formula for sine. Answer: 24/25 5. <E must be a right angle and so is <AOB. Since OA and OB are radii, triangle AOB is isosceles, so by SAS Similarity Theorem, the two triangle are similar triangles, so the ratio of the sides is 2, so the ratio of the areas is the square of the ratio or 2: Subdivide the hexagon into 6 triangles, using the center of the circle as a vertex for each of the six. Notice then, using the side of the hexagon as the base, the height of each triangle is the radius of circle A. So, the area can be found with 1 ap 2. Answer: Using the fact that the triangle is isosceles makes the altitude equal to 8. Use 1 ap 2 A to find the radius of the bottom circle, which is 3. Use similar triangles to find the radii in the middle and top circles. Answer: 3/ Extend BD to intersect the circle at E. Let DE = x. Use the Power of a Point to find x = 6. Extend OD to intersect the circle at F. We now have intersecting chords. Use the Power of a Point again. Let r = the radius. (r-2) * (r+2) = 6 * 3. Answer: r onstruct a triangle by connecting the center of the small circle to the point of tangency of the two circles to the midpoint of one side. Let r = radius of small circle (2 - r) 2 = (2 + r) 2 Answer: r = ½. 10. To find a locus, it helps to first try to find one or two points that satisfy the conditions. The two points of intersection have a power of 0, so they are two we might first see. Any other points, if they exist, must be somewhat centrally located. What about the points along the line that contains the two points of intersection of the circles? Proof?? 11. Using the Power of a Point, we know that TS = 8. Since PR and PW are radii, we can then use P of a P again to get: 64 x x 12. Answer: Draw ME and NE. You now have two right triangles that share a leg. Let x = AP. Run Pythagorean Theorem with triangles MEP and NEP. Answer: 5 2 5