Hello lance Originally Posted by
lance Give the equation of the circle both standard form and general form
Radius 3; tangent to x = -3 and y = 4 and below and to the left of these lines
Please help me and show the solution... Thanks!
I'm not entirely sure that I understand the question, but I'm assuming that the circle is to the left of $\displaystyle x = -3$ and below the line $\displaystyle y = 4$. If this is so, then its centre is where:
$\displaystyle x = -3-3=-6$ and $\displaystyle y = 4-3 = 1$
Now the equation with centre $\displaystyle (h,k)$ and radius $\displaystyle r$ has equation:
$\displaystyle (x-h)^2+(y-k)^2 = r^2$
so the equation of this circle is:
$\displaystyle (x+6)^2 + (y-1)^2 = 9$
and when we expand and simplify, this becomes:
$\displaystyle x^2+y^2+12x -2y -28=0$
Grandad