Question

Which is an equation of a circle with center (2, -1) that passes through the point (3, 4)?

(x - 2)2 + (y + 1)2 = 13
(x + 2)2 + (y - 1)2 = 13
(x + 2)2 + (y - 1)2 = 26
(x - 2)2 + (y + 1)2 = 26

Answer 1

Well, the form is (x-h)^2 + (y-k)^2 = r^2, where the minus does not change. So you be plugging in 2 and -1 into the form of (x-h) and (y-k). and the last part is r^2, which you can find by drawing a triangle using the center of the circle and your second point. If you draw a triangle and use pythagorean theorem, you can find the answer for r^2.

Answer 2

so r^2 is 26 right?

Answer 3

Looks right to me.

Answer 4

as mentioned above the general fporm of Equation of circle is
\[\Large (x-h)^2+(y-k)^2 =r^2\]
here (h,k) is the center of the circle.
since center of the circle is given (2,-1) it means h=2 and k=-1
put h=2 and k=-1 in the equation of the circle aftter plugging the values of h and k I am getting
\[\Large (x-2)^2+(y-(-1))^2=r^2\]
\[\Large (x-2)^2+(y+1)^2=r^2\]
we have two possible options which are similar . they are
\[\Large (x-2)^2+(y+1)^2=13\]
\[\Large (x-2)^2+(y+1)^2=26\]
to decide which one is correct put the given point (3,4) in the above two equation. the equation which will satify the point is the Amswer
plugging into first equation x=3 and y=4
\[\Large (3-2)^2+(4+1)^2=13\]
\[\Large 26 =13\] this is incorrect because 26 is not equal to 13!

plug x=3 and y=4 in the second equation

\[\Large (3-2)^2+(4+1)^2=26\]
\[\Large 26=26\] Yes this is Logical make sense!

so the correct Option should be
\[\Large (x-2)^2+(y+1)^2=26\]

Answer 5

@elizabethvilleda