Two circles have a radii of 15 and 95. If the two external tangents to the circles intersect at 60 degrees, how far apart are the centers of the circles?
Can someone please explain this to me, show the work and give me the answer. Thanks!
To find the distance between the centers of the circles, we can use the concept of similar triangles.
Let's assign labels to the different parts of the problem. Let O1 be the center of the circle with radius 15, O2 be the center of the circle with radius 95, and P be the point where the two external tangents intersect. Additionally, let A be the point where one of the tangents touches the circle with radius 15, and B be the point where one of the tangents touches the circle with radius 95.
First, let's draw a diagram to visualize the problem. Draw two circles with centers O1 and O2, and radii 15 and 95 respectively. Then, connect O1 and O2 with a segment. Next, draw two external tangents from O1 and O2 to intersect at point P. Finally, label the points of tangency as A (on the circle with radius 15) and B (on the circle with radius 95).
Now, let's consider the right-angled triangle O1AP. Since the line segment OA is perpendicular to the tangent at point A (by definition of a tangent), we have a right angle at O1. Angle O1AP is also 90 degrees minus 60 degrees (since the line segments OB and AP are parallel), which gives us an angle of 30 degrees. Thus, triangle O1AP is a right-angled triangle with angles of 30 degrees, 90 degrees, and 60 degrees.
By definition of a tangent, the line segment AP is tangent to the circle with radius 15, so the length of AP is equal to the radius 15.
Now, let's consider the right-angled triangle O1PB. Similar to triangle O1AP, we have a right angle at O1 and an angle of 30 degrees at O1PB. Since OP is the hypotenuse of both triangles, triangle O1PB is similar to triangle O1AP.
By ratio of sides in similar triangles, we can write:
(AP/O1P) = (OA/O1A)
Since AP is the radius 15, O1A is the radius 15, and OA is the distance between the centers of the circles, we can rewrite the equation as:
(15/O1P) = (OA/15)
Rearranging the equation, we get:
OA = (15/O1P) * 15
Now, let's consider the right-angled triangle O2PB. Like triangle O1PB, we have a right angle at O2 and an angle of 30 degrees at O2PB. Since OP is the hypotenuse of both triangles, triangle O2PB is also similar to triangle O1AP.
By ratio of sides in similar triangles, we can write:
(BP/O2P) = (BO2/O2B)
Since BP is the radius 95, O2B is the radius 95, and BO2 is the distance between the centers of the circles, we can rewrite the equation as:
(95/O2P) = (BO2/95)
Rearranging the equation, we get:
BO2 = (95/O2P) * 95
Since triangle O1PB and triangle O2PB share a common side PB, we can equate their ratios in terms of O1P and O2P:
(15/O1P) = (95/O2P)
Cross-multiplying, we get:
15 * O2P = 95 * O1P
Rearranging the equation, we get:
O2P = (95/O1P) * 15
Now, we can substitute the equations for OA and BO2 into the equation for O2P:
O2P = (95/O1P) * 15 = (95/(15/O2P)) * 15
Simplifying, we get:
O2P = (95/225) * 15 = (19/45) * 15 = (19/3)
Therefore, the distance between the centers of the circles is (19/3) units.