Respuesta :
The general equation of the circle is:
[tex](x-h)^{2} +(y-k)^{2} =r^{2} [/tex]
(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:
[tex](x+4)^{2} +(y+3)^{2} =r^{2} [/tex]
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:
[tex](6+4)^{2}+(2+3)^{2}=r^{2} \\ \\ 100+25=r^{2} [/tex]
Thus the value of r² is 125. Using the value in equation of the circle, we get:
[tex](x+4)^{2} +(y+3)^{2} =125 [/tex]
The radius of the circle is √125.
[tex](x-h)^{2} +(y-k)^{2} =r^{2} [/tex]
(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:
[tex](x+4)^{2} +(y+3)^{2} =r^{2} [/tex]
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:
[tex](6+4)^{2}+(2+3)^{2}=r^{2} \\ \\ 100+25=r^{2} [/tex]
Thus the value of r² is 125. Using the value in equation of the circle, we get:
[tex](x+4)^{2} +(y+3)^{2} =125 [/tex]
The radius of the circle is √125.
The equation for a circle in standard form is [tex](x-h) ^{2}+(y-k) ^{2} = r^{2} [/tex]. 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. [tex](6-(-4)) ^{2} +(2-(-3)) ^{2} = r^{2} [/tex]. Solving for [tex] r^{2} [/tex], we get that [tex] r^{2} =125[/tex], Now we can use the info given plus our newly found radius and write the equation. [tex](x+4) ^{2} +(y+3) ^{2} =125[/tex]