Question

Find the equation of circle with center at $( - 2,0)$ and radius equal to $5$ units and also plot it on a graph.

Answer 1

In this question we need to find the equation of circle through a given point and of specified radius. This type of question can be solved easily, if the student remembers the standard equation of the circle. By substituting the values in the standard equation and then simplifying it, answers can be obtained.

Complete step by step answer:
Let the given center for the circle be A $( - 2,0)$ and radius be r $= 5$ units.
The standard equation of circle is given by the formula : ${r^2} = {(x - h)^2} + {(y - k)^2}$
Where, h and k are the x and y-coordinate of the center of the circle respectively and r is the radius. Now, we will substitute the values in the standard equation to get;

\Rightarrow 25 = {x^2} + 4 + 4x + {y^2} \\\ \therefore {x^2} + {y^2} + 4x = 21 \\\ $$ The equation obtained above is the required equation of circle through A $( - 2,0)$ and radius r $ = 5$ units. Now, to plot this circle in graph, mark point A $( - 2,0)$ and draw a circle of radius $5$ units. Plot of the circle is as shown in the diagram.  **Note:** Students are advised to memorize the standard equation of circle. This comes handy during solving problems in competitive exams. Also, there is one general form of equation of circle, which is: $${x^2}\; + \;{y^2}\; + \;2gx\; + \;2fy\; + \;c\; = \;0$$ , where the center of circle is represented by $$\;( - g, - f)$$ and its radius, $r$ is given by $$\;{r^2}\; = \;{g^{2\;}} + \;{f^2} - \;c$$ . If the radius of the circle, $${g^{2\;}} + \;{f^2} - \;c > 0$$ , then the general equation of the circle represents a real circle. If the radius of circle, $$$$$${g^{2\;}} + \;{f^2} - \;c < 0$$, then the general equation of circle represents a imaginary circle; and if the radius of circle, $$$$$${g^{2\;}} + \;{f^2} - \;c = 0$$ , then the circle represented is a point circle. Students must remember the general form also.