Question

What is the equation of the circle with a center at $\left( -1,-1 \right)$ and a radius of 8?

Answer 1

Hint: The general equation of a circle in the Cartesian plane or the X-Y coordinate axis can be used to solve this problem. Assuming $\left( {{x}_{1}},{{y}_{1}} \right)$ to be the center of the circle and A be the radius, then the equation of circle is given by, ${{\left( x-{{x}_{1}} \right)}^{2}}+{{\left( y-{{y}_{1}} \right)}^{2}}={{A}^{2}}$ . Use this formula to solve this problem.

Complete step by step answer:
In the question, we are given both the center and radius of the circle. Centre is given as $\left( -1,-1 \right)$ and the radius is given as 8. Hence, we can use these values and substitute them in the general equation of a circle in the Cartesian plane to find its equation. Every point on the circumference of the circle will be equidistant from the center of the circle. The circle is as shown on the Cartesian plane in the figure.

The general equation of a circle is given by ${{\left( x-{{x}_{1}} \right)}^{2}}+{{\left( y-{{y}_{1}} \right)}^{2}}={{A}^{2}}$, where $\left( {{x}_{1}},{{y}_{1}} \right)$ is the center of the circle and A is the radius of the circle. Substituting the values given to us in the question, we get,
\[\begin{align}
  & {{\left( x-\left( -1 \right) \right)}^{2}}+{{\left( y-\left( -1 \right) \right)}^{2}}={{8}^{2}} \\
 & \Rightarrow {{\left( x+1 \right)}^{2}}+{{\left( y+1 \right)}^{2}}=64 \\
\end{align}\]
Therefore, the required equation of the circle is found to be \[{{\left( x+1 \right)}^{2}}+{{\left( y+1 \right)}^{2}}=64\].

Note: While solving this problem, one has to be careful in taking the positive and negative sign. If the center has negative coordinates, then the equation will have positive terms like the one we have solved. If the given coordinates are positive, then in the equation, they will become negative. Hence, one has to be careful in taking the sign of the coordinate given. Any small change in the sign will represent a circle other than the one which is required.