Solve Equation of a Circle: Get Help Now!

[odolwa99]

I'm having some difficulty with this question. Can anyone help me out, please?

Many thanks.

Homework Statement

A circle of radius length $\sqrt{20}$ contains the point (-1, 3). Its centre lies on the line x + y = 0. Find the equations of the 2 circles that satisfy these conditions.

Homework Equations

The Attempt at a Solution

Circle equation for the point: $(-1-h)^2+(3-k)^2=20\rightarrow h^2+k^2+2h-6k-10=0$
Because x+y=0 contains centre c as (-g, -f), we can say -g - f = 0.
Thus, -g = -2h & -f = 6k.
And so, $x+y=0\rightarrow -2h+6k=0\rightarrow h=3k$.
Returning to circle equation: $(3k)^2+k^2+2(3k)-6k-10=0\rightarrow 10k^2=10\rightarrow k=+/- 1$
For k = 1: $h=3k\rightarrow h=3$.
Thus, centre c = (3, 1) and equation = $(x-3)^2+(y-1)^2=20\rightarrow x^2+y^2-6x-2y-10=0$
For k = -1: $h=-3$
Thus, centre c = (-3, -1) and equation = $(x+3)^2+(y+1)^2=20\rightarrow x^2+y^2+6x+2y-10=0$

Ans.: (From textbook): $x^2+y^2-2x+2y-18=0$ & $x^2+y^2+10x-10y+30=0$

 

[CAF123]

You can't have a centre of (3,1) for example because the centre must lie on the line x=-y, so in your notation the centre is like (h,-h). The condition you want is that the distance between the centre and the point is less or equal to $\sqrt{20}$. That is, $$\sqrt{(h+1)^2 + (k-3)^2} \leq \sqrt{20}$$ Use this and the condtion that h=-k, h and k denoting the coordinates of the centre.

EDIT: I misinterpreted what 'contained' meant - I think it should be that the point lies on the circumference of the circle so there is a strict equality in the above expression.

 

[odolwa99]

Ok, I have it now. Thank you very much.