Question 37.
There is a circle with center at point O. From an external point P, a tangent PT is drawn at point T of the circle. From P, a secant PR is drawn which intersects the circle at points J and R such that J is between P and R. There is a point N between J and R. JR is not a diameter.
PT = 12, PJ = 8, ON = 4, NJ = 6. What is the radius of the circle?
Answer 37.
PR * PJ = PT^2
=> PR = (12)^2 / 8 = 18
=> JR = 18 - 8 = 10
Let M = midpoint of JR,
then JM = 5 and MN = NJ - JM = 6 - 5 = 1
Radius^2
= JM^2 + OM^2
= JM^2 + ON^2 - MN^2
= (5)^2 + (4)^2 - 1^2
= 40
=> Radius
= √(40)
= 2√(10).
LINK to YA!
There is a circle with center at point O. From an external point P, a tangent PT is drawn at point T of the circle. From P, a secant PR is drawn which intersects the circle at points J and R such that J is between P and R. There is a point N between J and R. JR is not a diameter.
PT = 12, PJ = 8, ON = 4, NJ = 6. What is the radius of the circle?
Answer 37.
PR * PJ = PT^2
=> PR = (12)^2 / 8 = 18
=> JR = 18 - 8 = 10
Let M = midpoint of JR,
then JM = 5 and MN = NJ - JM = 6 - 5 = 1
Radius^2
= JM^2 + OM^2
= JM^2 + ON^2 - MN^2
= (5)^2 + (4)^2 - 1^2
= 40
=> Radius
= √(40)
= 2√(10).
LINK to YA!
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