Question

Write the standard form of the equation of the circle shown.

Answer 1

The stand form of the circle is

$(x-h)^{2} + (y-k)^{2} = r^{2}$

here (h,k) is the center of the cirlce and (x,y) is the point on the circle.

In our case (h,k) is (-2,2)

so our circle equation is

$(x--2)^{2} + (y-2)^{2} = r^{2}$.      ----------- (i)

and we can find our r by finding the distance between center and the point given on the circle. lets say we take (1,0) point given on the circle according to our diagram

Then

$r = \sqrt{(-2-1)^{2}+(2-0)^{2}}$

$r = \sqrt{9+4}$

$r = \sqrt{13}$

so by putting the values of r in equation (i)

$(x--2)^{2} + (y-2)^{2} = \sqrt{13} ^{2}$

$(x+2)^{2} + (y-2)^{2} = 13$

Which is our third option.

Answer 2

Answer:

$(x+2)^2+(y-2)^2} =13$

Step-by-step explanation:

We are given a circle with a center at the point (4, -5) and two points on the circle (-5, 0) and (1, 0).

We know that the standard form of equation of a circle is:

$r=\sqrt{(x_1-x)^2+(y_1-y)^2}$

So putting the given values to get the value of radius (r):

$r=\sqrt{(-2-1)^2+(2-0)^2}$

$r=\sqrt{13}$

Substituting the value of radius to get the equation of the circle:

$\sqrt{(x+2)^2+(y-2)^2} =\sqrt{13}$

Taking square on both the sides:

$(x+2)^2+(y-2)^2} =13$

Therefore, the equation of the given circle is $(x+2)^2+(y-2)^2} =13$.