Question

The value of p for which the circles x 2 + y 2 – px + 3y – 7=0 and x 2 + y 2 – 16x – 9py – 4 = 0 cut orthogonally is _

Answer 1

Answer:

$\text{Thevalue of p is }-\frac{6}{11}$

Step-by-step explanation:

Given the two circles

$x^2+y^2-px+3y-7=0$ and $x^2+y^2-16x-9py-4=0$

The equation for the two given circles whose equations are

$x^2 + y^2 + 2gx + 2fy + c = 0$ and $x^2 + y^2 + 2g'x + 2f'y + c' = 0$, to be orthogonal is given by:

$2gg' +2ff'=c+c'$

Comparing given equations with standard equations, we get

2g=-p, 2f=3, c=-7

g'=-8, 2f'=-9p, c'=-4

Hence, if the given two circle cuts orthogonally then

$(-p)(-8)+3(\frac{-9p}{2})=7-4$

⇒ $8p-\frac{27p}{2}=3$

⇒ $16p-27p=6$

⇒ $-11p=6$

⇒ $p=-\frac{6}{11}$