Question

PLEASE I NEED HELP

Answer 1

The general equation of the circle is:

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

(h,k) is the center of the circle and r is the radius. We are given the coordinates of the center. So we can write the equation of the circle as:

$(x+4)^{2} +(y+3)^{2} =r^{2}$

In order to complete the equation, we need the radius of the circle. Using the given point we can find the radius. Using the point in the above equation, we get:

$(6+4)^{2}+(2+3)^{2}=r^{2} \\ \\ 100+25=r^{2}$

Thus the value of r² is 125. Using the value in equation of the circle, we get:

$(x+4)^{2} +(y+3)^{2} =125$

The radius of the circle is √125.

Answer 2

The equation for a circle in standard form is $(x-h) ^{2}+(y-k) ^{2} = r^{2}$.  We have the center, so we have the h and the k, and we have a point, so we have the x and the y.  Let's fill in accordingly and simplify. $(6-(-4)) ^{2} +(2-(-3)) ^{2} = r^{2}$.  Solving for $r^{2}$, we get that $r^{2} =125$,  Now we can use the info given plus our newly found radius and write the equation.  $(x+4) ^{2} +(y+3) ^{2} =125$