Hi, I have this equation: x^2+12x+y^2-6y+29=0 and need to find it's centre and radius, also the coordinates of any points at which the circle intersects the line y = x+13. I have had a few stabs at finding the centre and radius but think I go wrong at one partcular point so could someone perhaps explain the method so I can clarify in my mind the correct precedure. Thanks.

 

The general equation for a circle is:
(x-a)^2 + (y-b)^2 = r^2


the coordinate (a,b) will be the center and r will be the radius.

to make sense of this, make a perfect square with x and y, and then you have the center point and whatever is left over is the radius squared. Then to solve for intersections, just solve the system of equations.

In this case subtract the 29 over to get

x^2 + 12x +y^2 - 6y = -29

then make a perfect square

x^2 + 12x + A + y^2 -6y + B = -29 + A + B

=> x^2 + 12x + A = (x + something) ^2
(same thing for y)
Solve for A and B (ie find a value of A and B to make the equation look like the general equation for a circle), and you have the equation of the circle, the rest is easy.

 

Using DrNick's hints take a stab at the solution and then post your efforts here.

We can then give you a hand with anything that trips you up.

hgmjr

 

all i can say is above problem can be solved with diff approach
in fact there is a direct formula but you must derive it first to get a better feel
of things try reading abt general equation for conic sections