1. Two circles intersect and have a common chord. One of the circles has a radius of 10 cm and the other a radius of 17cm. If their centers are 21 cm apart, what is the length of the common chord.
16cm
Let x = ½ the length of the
common chord. Let y = part of the segment joining the circle centers
and 21 – y = the other part of the segment. Use the Pythagorean
theorem on each right triangle.
Isolate
in the first equation
. Replace
in the 2nd equation with
. 
Solve for y.
Once
you know y = 6 use the pythagorean theorem to solve for x.
2. An isosceles triangle with each leg measuring 25 units is inscribed in a circle. If the height of the triangle is 7 units, find the radius of the circle.
44.6 cm 
Draw
in the extra radius and label each of the missing segment length.
Note that radii of a circle are congruent and that is why the introduced
radii has y + 7 on it. Use the pythag. theorem on the upper right
triangle. 
Once
you solve for x use the bottom right triangle and repeat the pythag.
theorem. 
The radius
= y+7 so 37.6+7 =44.6
3. Given
circle O with
, BC = 5 and PQ = 30, find the radius of the circle.
25
Introduce
PO as a radius. Label CO, x. Since OB and PO are both radii of
a circle each has the same measure of x+5. A radius drawn
perpendicular to a chord bisects the chord so since PQ= 30, PC= 15.
Set up a pythag on the right triangle. 
4. Rectangle
RECT is inscribed in a circle. The 
. If
CT = 12 cm, find the area of the circle
Since
the arc is 240 the remaining portion of the circle is 120. <ECR
is inscribed on the 120 degree arc. Introduce the diagonal which is
the diameter of the circle. Since an inscribed angle is 1/2 the measure of
its intercepted arc, it is a 60 degee angle. If CT =12, the opposite sides
in rectangles are congruent so ER = 12. Use the ratios of a 30-60-90
triangle to find the short leg and hypotenuse. Remember to get back
from the long leg you must divide by 
This makes the short leg 
.
The radius is 1/2 the diameter which is 
48p
5. Given circles A, B, and C are externally tangent. If AB = 15, AC = 12 and BC = 11, find the radius of circle B.
This
problem is a walkaround. Let x be the radius you are looking
for. Then assign values around the circle as shown. Note that
on side AC 11-x + 15 - x = 12. Solve this equation by
combining like terms 26 - 2x = 12, subracting 26 from both
sides, -2x = -14 and dividing by -2, x = 7
7
6. Given
circle A with tangent
. If AB = 9 and CD = 6, find
the length of
.
12 Since AB = 9 so does AD as they are both radii of the circle. Adding AD + DC = 15 . Radii are perpendicular to the tangents. Use pythag to solve for BC.
7. If
is inscribed in circle O,
then O is the point of intersection of the _?_of
.
a) altitudes b) medians c) angle bisectors d) ^ bisectors of the sides
8. The length of a side of an equilateral triangle is 6. Find the length of the radius of the circle inscribed in the triangle.
Draw in the radius and the segment to the corner angle. This forms a
30-60-90 triangle. Since the tangent point is at the midpoint of the
triangle when the triangle is equilateral the long leg of the 30-60-90 =
3
a)
b)
c)
d) 3
9. In
circle S if
and
is tangent to the circle,
then the m
If the arc = 105 then the central angle is 105. Then <VSU = 75 because it is a supplement to the 105 angle. There is already a right angle due to the radius drawn to the point of tangency. The three angles of a triangle add to 180 - subtract 165 and the result is 15 degrees.
a) 15° b) 30° c) 37.5° d) 75°
10. Circle
O and P have radii of 15 cm and 8 cm.
Their center are 25 cm apart.
Find the length of the common external tangent segment
.
24 cm Draw in a perpendicular from P
to segment QO. This will form a rectangle. The opposite sides
of a rectangle are congruent so mark as 8. Since the radius is 15
there are 7 units left on the radius which makes up a leg of a right
triangle. The distance between the circle centers is 25 which
is the hypotenuse of the right triangle. Use the pythag. theorem and find
the missing side. This will be the same length as QR,
