Find the eq. of the circle which passes through (1,1) and cuts orthogonally x2 + y2-8x-2y+16=0 and x2+y2-4x-4y-1=0
Previously i did a similar kind of sum for a circle which passed through (0,0) so there was no constant term involved but when it passes through (1,1) there is the f term , g term and c the constant and so I am confused
Thank You!
1Take the standard form of the circle
x2 + y2 + 2gx + 2fy + c = 0
It passes through (1,1). So we get one equation. Since the required circle is orthogonal to the two given circles, apply condition for orthogonal circles. So we get two more equations. Hence we have three equations in total and three unknowns i.e g,f and c. We can calculate them by solving the three equations.
3That was exactly what I did but then I was getting the wrong answer :(
3I am getting g as -7/3
c=-5 and f=23/6 and so i got the eq. 6x2+6y2-28x+23y-30=0
while the answer is 3x2+3y2-14x+23y-15=0...he is getting the lcm as 3
1you got the correct g, f and c values. But you substituted them wrongly in the equation.
you'll get
x2 + y2 - 143x + 233y - 5 = 0
Multiply the equation by 3 and you get the required equation.