Question

Find the equation of the circle with center at (3, 2) and through the point (5, 4).

Answer 1

\(\large\color{black}{ (x-h)^2+(y-k)^2=r^2 }\)

where (h,k) is the cementer

Answer 2

so to figure out the left side, is not very hard, (right ?)

Answer 3

for the right side, radius is a distance from the center to any point on the circle.
In your case the distance from (3,2) to (5,4)

Answer 4

can you find the distance between these 2 points ?

Answer 5

oh, my first reply should say that (h,k) is the center of the circle.

Answer 6

oh ok so (x-3)^2+(y-2)^2=r2

Answer 7

yes, that left side is correct, and now you have to find r^2

Answer 8

do you know the distance formula , or should I tell you it?

Answer 9

nope,so many formulas to remember I forgot

Answer 10

\(\Large\color{black}{ \displaystyle {\rm D} =\sqrt{ (\color{blue}{{\rm y}_1}-\color{red}{{\rm y}_2})^2+(\color{green}{{\rm x}_1}-\color{darkgoldenrod}{{\rm x}_2})^2 }}\)


where
\(\Large\color{black}{ \displaystyle {\rm D} }\) is the distance.

\(\Large\color{black}{ \displaystyle (\color{green}{{\rm x}_1}~,~~\color{blue}{{\rm y}_1}) }\) and \(\Large\color{black}{ \displaystyle (\color{darkgoldenrod}{{\rm x}_2}~,~~\color{red}{{\rm y}_2}) }\) are your two points.


(Hoe this makes sense)

Answer 11

your two points are (as you have probably noticed already) are
(3,2) and (5,4)