center at (23,-1) externally tangent to x^2+y^2-34x-14y+313=0
How do you do this if the answer is in standard form. P.s. I just don't know how to get the radius.
2026-03-28 01:02:54.1774659774
Equation of a circle problem
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The equation of a circle with center $A$,$B$ and radius $r$ is:
$(x-a)^2 + (Y-b)^2 = r^2$
The equation of the circle that you start with is $x^2 + y^2 +-34x - 14y +313 = 0$.
So you rearrange this to give $(x-17)^2 + (y-7)^2 - 289 - 49 + 313 = 0$,
which simplifies to $(x-17)^2 + (y-7)^2 = 25$.
Taking the generic equation of a circle, this shows that $r^2 = 25$, therefore radius $r= 5$.