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Consider the circle x^2+y^2 -8x-18y +93...

Consider the circle `x^2+y^2 -8x-18y +93=0` with the center C and a point `P(2,5)` out side it. From P a pair of tangents PQ and PR are drawn to the circle with S as mid point of QR. The line joining P to C intersects the given circle at A and B. Which of the following hold (s)

A

CP is the arithmetic mean of AP and BP

B

PR is the geometric mean of PS and PC

C

PS is the harmonic mean of PA and PB

D

The angle between the two tangents from P is `tan^(-1)((4)/(3))`

Text Solution

Verified by Experts

The correct Answer is:
A, B, C, D

Radius `= sqrt(16 +81-93) = 2`
`CP = sqrt(20), AP = sqrt(20) -2, BP = sqrt(20) +2`
`rArr CP = (AP +BP)/(2)`
Thus, CP is the arithmetic mean of AP and BP
Now let, `L = PR = sqrt((PC)^(2)-r^(2)) = sqrt(20-4) = 4 = PQ`
`:. tan theta = (2)/(4) =(1)/(2)`
Also, `cos theta = (PS)/(PR)`
`rArr PS = PR cos theta 4 .((2)/(sqrt(5))) = (8)/(sqrt(5))`
Harmonic mean between PA and PB
`= (2(sqrt(20)-2)(sqrt(20+2)))/(2sqrt(20)) = (16)/(2sqrt(5)) = (8)/(sqrt(8)) =PS`
Thus, PS is the harmonic mean of PA and PB.
`(PS) (CP) = ((8)/(sqrt(5)))(sqrt(20))=16=(PR)^(2)`
Thus, PR is the geometric mean of PS and CP.
Now, angle between the two tangents.
`= 2 theta = 2 tan^(-1) ((1)/(2)) = tan^(-1) ((2.(1)/(2))/(1-(1)/(4)))=tan^(-1)((4)/(3))`
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