Question

What is the equation of a circle that passes through (2, 3) and has its center at Q(2, -1)?

Answer 1

\[(x-2)^2+(y-3)^2=(2-2)^2+4^2\]

Answer 2

If you still don't undertand...I will help you !

Answer 3

Please help

Answer 4

i surely don't

Answer 5

circle with radius \(r\) and center \((h,k)\) had the equation
\[(x-h)^2+(y-k)^2=r^2\]
in your case you have the center at \((2,-1)\) and so \(h=2,k=-1\)

Answer 6

replace \(h\) by 2, \(k\) by -1 and get \((x-2)^2+(y+1)^2=r^2\) all that is left is to find \(r^2\)

Answer 7

okk..you know "equation of a circle" ? by knowing his center and radius ?!

Answer 8

the radius is the distance between the center and any point on the circle
the distance between \((2,3)\) and \((2,-1)\) is 4 since they have the same first coordinate

Answer 9

Is the answer, 16?

Answer 10

so finish via \[(x-2)^2+(y+1)^2=4^2\]

Answer 11

yes, 16 on the right

Answer 12

Thank you!:D

Answer 13

@ satellite73 thank you you saved me !