S≡x2+y2−6x+8y+k=0 ⇒(x−3)2+(y+4)2=25−k Thus, the center of the circle C is at (3,−4)=(x1,y1). Let point P as (x2,y2) and point Q is the point that divides the line segment CP in the ratio −1:2 Also, 3x+4y−43=0 ⇒y=
−3
4
x+
43
3
So, slope of tangent line =
−3
4
And slope of line CP=−(
−4
3
)=
4
3
Thus, equation of line CP is ⇒y−(−4)=
4
3
(x−3) ⇒y+4=
4
3
(x−3) ⇒y=
4
3
x−8 Since, P lies on both the lines 3x+4y−43=0 and the line CP, ∴3x+4(
4
3
x−8)−43=0 ⇒3x+
16
3
x−32−43=0 ⇒
25
3
x=75 ⇒x=9 Now, y=
4
3
x−8=
4
3
(9)−8 =12−8=4 So, P=(9,4) Using section formula, we find the coordinates of Q(x,y). x=
2x1−x2
2−1
=
2(3)−9
1
=−3 y=
2y1−y2
2−1
=
2(−4)−4
1
=−12 So, Q=(x,y)=(−3,−12) The power of point Q w.r.t to circle, x2+y2−6x+8y+k=0 is (−3)2+(−12)2−6(−3)+8(−12)+k =9+144+18−96+k =75+k⋅⋅⋅⋅⋅⋅⋅(i) Now, the line 3x+4y−43=0 is tangent to the circle, the radius of the circle is the distance from the center C(3,−4) to the line 3x+4y−43=0 So, r=
|Ax0+By0+C|
√A2+B2
=
|3(3)+4(−4)−43|
√32+42
=
|9−16−43|
5
=
50
5
=10 Now, r=√25−k ⇒10=√25−k⇒25−k=100 ⇒k=−75 Put k=−75 in Eq. (i), we get The power of Q w.r.t the circle =75+k=75−75=0
