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)

Options

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))`

Answer

A, B, C, D

Explanation

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))`