Given equation of circle is C=x2+y2−16x−12y+64=0 ......(i) (i) Equation of polar at (−5,1) w.r.t. to C is x(−5)+y(1)−8(x−5)−6(y+1)+64=0 ⇒ −5x+y−8x+40−6y−6+64=0 ⇒ −13x−5y+98=0 ⇒ 13x+5y=98 (ii) On differentiating Eq. (i) w.r.t. to x, we get 2x+2y
dy
dx
−16−12
dy
dx
+0=0 ⇒ (2y−12)
dy
dx
=(16−2x) ⇒
dy
dx
=(
8−x
y−6
) At (8,0)(
dy
dx
)(8,0)=
8−8
0−6
=0 ∴ Equation of tangent at (8,0) is (y−0)=0(x−8) ⇒ y = 0 (iii) Slope of normal is
dy
dx
=
6−y
8−x
(
dy
dx
)(2,6)=
6−6
8−2
=0 Equation of normal is (y−6)=0(x−2) ⇒ y=6 (iv) Equation of the diameter of circle through ( 8 ,12 ) is x=8