pre calc word problem

Question

a circle C has center at the origin and radius 7, another circle K has a diameter with one end at the origin and the other end at the point (0,17) the circles C and K intersect in two points. let P be the point of intersection of C and K which lies in the first quadrant. let (r,∅) be the polar coordinates of P, chosen so that r is positive and 0≤∅≤2 find r and ∅

Answer 1

Circle C

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x^2 + y^2 = 49

circle K <--- not the gas station...lol ;-)

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diameter (0,0) and (0,17)....

center is at (0,8.5)

r = 8.5

x^2 + (y-8.5)^2 = 72.25

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intersection:

x^2 + y^2 = 49

x^2 + (y-8.5)^2 = 72.25

Subtracts the equations:

(y-8.5)^2 - y^2 = 72.25 - 49

y^2 - 17y + 72.25 - y^2 = 23.25

-17y + 72.25 = 23.25

-17y = -49

y = 49/17

x^2 + (-49/17)^2 = 49

x^2 + 2401/289 = 49

x^2 = (49*289 - 2401)/289

x^2 = 49(289 - 49)/289

= 49 ( 240)/289

= 49 ( 24*10) / 289

= 49( 2*2*2*3*2*5)/ 289

so x = +or- 7 * 2*2 * sqrt(15)/ 17

= +or- (28/17) sqrt(15)

Polar, in quadrant 1:

r^2 = x^2 + y^2, x=(28/17) sqrt(15); y = 49/17

x^2 + y^2 = [28^2 * 15 + 49^2]/289

= [ (7*4)^2 * 15 + (7*7)^2]/ 289

= 49[ 16*15 + 49]/289

= 49[ 240 + 49]/289

= 49[289]/289

= 49 <--- which shows the answer is correct

x = r cos T ---> (28/17) sqrt(15) = 7 * cos T

(4/17) sqrt(15) = cos T

T = arccos [(4/17) sqrt(15)] =

= 24.315739....

Note that 7* sin T is approximately = 2.882352....

which is (approximately) y

Answer 2

This is more an exercise in computational skills than it is in Trigonometry.

x² + y² = 49

(x - 8.5)² + y² = 8.5²

Substituting into y²

x² - 17x + 8.5² + 49 - x² = 8.5²

-17x + 49 = 0

49 = 17x

49 / 17 = x

y² = 49 - (49 / 17)²

Combining using LCD 17²

y² = 11760 / 17²

y = 28 √15 / 17

P = (49 / 17, 28√15 / 17)

Leave it to you to put into polar coordinates.