Theequation of circle with center at \(\left( { - 3,12} \right)\) and radius \(5{\text{ units}}\) is \(\boxed{{x^2} + {y^2} + 6x - 24y + 128 = 0}.\)
Further explanation:
The standard equation of the circle with center \(\left( {h,k} \right)\) and radius r can be expressed as,
\({\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}.\)
Given:
A circle with center at \(\left( { - 3,12} \right).\)
Radius of the circle is \(5{\text{ units}}.\)
Explanation:
The center is at \(\left( { - 3,12}\right).\)
Compare the general form \(\left( {h,k} \right)\) of the center to the given center \(\left( { - 3,12} \right)\) to obtain the value of \(h\) and \(k\).
Therefore, the value of \(h\) and \(k\) are \(-3\) and \(12\) respectively.
The radius of the circle is \(5{\text{ units}}.\)
Substitute \(- 3\) for \(h\), \(12\) for \(k\) and \(5\) for \(r\) to obtain the standard equation of the circle.
\(\begin{aligned}{\left( {x - \left( { - 3} \right)} \right)^2} + {\left({y - 12} \right)^2}&= {\left( 5 \right)^2}\\{\left( {x + 3} \right)^2} + {\left( {y - 12}\right)^2} &= 25\\{x^2} + 6x + 9 + {y^2} - 24y + 144&= 25\\{x^2} + {y^2} + 6x - 24y + 153&= 25\\{x^2} + {y^2} + 6x - 24y + 153 - 25&= 0\\{x^2} + {y^2} + 6x - 24y + 128&= 0\\\end{aligned}\)
The equation of circle with center at \(\left( { - 3,12}\right)\) and radius \(5{\text{ units}}\) is \(\boxed{{x^2} + {y^2} + 6x - 24y + 128 = 0}.\)
Learn more:
1. Learn more about equation of circle brainly.com/question/1506955.
2. Learn more about domain of the function https://brainly.com/question/3852778.
3. Learn more about coplanar https://brainly.com/question/4165000.
Answer details:
Grade: Middle School
Subject: Mathematics
Chapter: Circle
Keywords: Circle, standard form of the circle, equation of the circle, center, diameter of circle, radius of the circle, center A.
