The equation of the circle is [tex]\rm (x+6)^2+(y-5)^2=29[/tex].
We have to write an equation for a circle that has a center at (-6,5) and passes through the point (-11,3).
According to the question,
To determine the equation of the circle following all the steps given below.
The standard form of the equation of the circle is,
[tex]\rm (x-a)^2+(y-b)^2=r^2[/tex]
Where a and b are the center of the circle.
The equation of the circle which passes through the center is (-6, 5) is,
[tex]\rm (x-(-6))^2+(y-5)^2=r^2\\\\\rm (x+6)^2+(y-5)^2=r^2[/tex]
And the circle passes through the point (-11, 3) is,
[tex]\rm (x+6)^2+(y-5)^2=r^2\\\\ (-11+6)^2+(3-5)^2=r^2\\\\(-5)^2+(-2)^2=r^2\\\\25+4=r^2\\\\r^2=29\\[/tex]
Therefore,
The equation of the circle which passes through the center is (-6, 5) is,
[tex]\rm (x+6)^2+(y-5)^2=29[/tex]
Hence, the required equation of the circle is [tex]\rm (x+6)^2+(y-5)^2=29[/tex].
To know more about Circles click the link given below.
https://brainly.com/question/9549018
