Given equation of circle is x2+y2−2x+10y−38=0. (A) Polar equation at point (4,3) is S1=0. ⇒x×4+y×3−(x+4)+5(y+3)−38=0 ⇒3x+8y=27 (B) Equation of tangent at point (9,−5) is x×9+y×(−5)−(x+9)+5(y−5)−38=0 ⇒ 8x−72=0 ⇒ x=9 (C) On differentiating given equation w.r.t. x, we get 2x+2y
dy
dx
−2+10
dy
dx
−0=0 ⇒
dy
dx
(2y+10)=2−2x ⇒ (
dy
dx
)(−7,−5)=
2−2×(−7)
2×(−5)+10
=
16
0
=
1
0
∴ Equation of normal at point (−7,−5) is y+5=−0(x+7) ⇒ y+5=0 (D) Centre of circle C is (1,−5). ∴ Equation of diameter passing through (1,−5) and (1,3) is y+5=