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Question: How do you write the equation for a circle with center \(\left( -2,-3 \right)\) and \(r=3?\)...

How do you write the equation for a circle with center (−2,−3)\left( -2,-3 \right) and r=3?r=3?

Explanation

Solution

We will use the standard equation of the circle to find the equation of the given circle. Suppose that we are given a circle centered at the point (h,k)\left( h,k \right) with a radius r.r. The equation of this circle is given by the equation (x−h)2+(y−k)2=r2.{{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}}.

Complete step by step solution:
Let us consider the given circle centered at the point (−2,−3)\left( -2,-3 \right) with a radius of r=3.r=3.
We are asked to write the equation for this circle.
For that, we need to use the standard equation of a circle.
If we have a circle that is centered at a point (h,k)\left( h,k \right) with a radius of r,r, then the equation of the circle is given by the equation (x−h)2+(y−k)2=r2.{{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}}.
The above written equation is the standard equation of a circle centered at a point (h,k)\left( h,k \right) with a radius of r.r.
Let us use the standard equation to write the equation of the given circle.
We need to compare the values.
So, we will obtain the center (h,k)=(−2,−3).\left( h,k \right)=\left( -2,-3 \right).
From this, we will get the x−x-coordinate and the y−y-coordinate of the given circle.
So, we will get h=−2h=-2 and k=−3.k=-3.
Also, from the question, we have r=3.r=3.
Therefore, the equation will become (x−(−2))2+(y−(−3))2=32.{{\left( x-\left( -2 \right) \right)}^{2}}+{{\left( y-\left( -3 \right) \right)}^{2}}={{3}^{2}}.
Now, we will get this equation as (x+2)2+(y+3)2=9.{{\left( x+2 \right)}^{2}}+{{\left( y+3 \right)}^{2}}=9.
Let us expand the square terms to get x2+4x+4+y2+6y+9=9.{{x}^{2}}+4x+4+{{y}^{2}}+6y+9=9.
Let us cancel 99 from both sides to get x2+4x+y2+6y+4=0.{{x}^{2}}+4x+{{y}^{2}}+6y+4=0.
Hence the equation of the given circle is x2+y2+4x+6y+4=0.{{x}^{2}}+{{y}^{2}}+4x+6y+4=0.

Note: Remember the identity we have used in the problem to expand the square terms (a+b)2=a2+2ab+b2.{{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}. The equation of the circle centered at origin (0,0)\left( 0,0 \right) with a radius of rr is given by x2+y2=r2.{{x}^{2}}+{{y}^{2}}={{r}^{2}}.