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Question: How do you write the standard form of the equation of the circle with center \(\left( 3,-2 \right)\)...

How do you write the standard form of the equation of the circle with center (3,−2)\left( 3,-2 \right) Radius 33?

Explanation

Solution

In this problem we have to find standard form of the equation of the circle by using the center of circle with radius .the standard form of circle is (x−x1)+(y−y1)=r2\left( x-{{x}_{1}} \right)+\left( y-{{y}_{1}} \right)={{r}^{2}}, xx and yy are the variables ,x1{{x}_{1}} is the x−coordinatex-coordinate of the center, y1{{y}_{1}} is the y−coordinatey-coordinate of the center and rr is the radius of the circle, so in question they have given center i.e., we have the values of x1{{x}_{1}} ,y1{{y}_{1}} and also they have mentioned the radius rr. Now substituting the values in standard form of the circle equation then will we get the result.

Formula Used:
The standard form of the equation of the circle is
(x−x1)2+(y−y1)2=r2{{\left( x-{{x}_{1}} \right)}^{2}}+{{\left( y-{{y}_{1}} \right)}^{2}}={{r}^{2}}

Complete Step by Step Procedure:
Given that,
The center of circle is (3,−2)\left( 3,-2 \right), and radius r=3r=3
From the center of the circle the values of x1{{x}_{1}}, y1{{y}_{1}} are
x1=3 y1=−2 \begin{aligned} & {{x}_{1}}=3 \\\ & {{y}_{1}}=-2 \\\ \end{aligned}
We know that the standard form of the equation of the circle
(x−x1)2+(y−y1)2=r2{{\left( x-{{x}_{1}} \right)}^{2}}+{{\left( y-{{y}_{1}} \right)}^{2}}={{r}^{2}}
Now we will substitute center values in the above equation, then we will get
(x−3)2+(y−(−2))2=r2{{\left( x-3 \right)}^{2}}+{{\left( y-\left( -2 \right) \right)}^{2}}={{r}^{2}}
Now simplify the above equation, then
(x−3)2+(y+2)2=r2{{\left( x-3 \right)}^{2}}+{{\left( y+2 \right)}^{2}}={{r}^{2}}
Now we will replace rrwith radius of the circle
(x−3)2+(y+2)2=(3)2{{\left( x-3 \right)}^{2}}+{{\left( y+2 \right)}^{2}}={{(3)}^{2}}
We know that 32=9{{3}^{2}}=9, then we will get
(x−3)2+(y+2)2=9{{\left( x-3 \right)}^{2}}+{{\left( y+2 \right)}^{2}}=9

Hence the standard from of the circle is
∴(x−3)2+(y+2)2=9\therefore {{\left( x-3 \right)}^{2}}+{{\left( y+2 \right)}^{2}}=9

Note:
In the question they have only asked to calculate the standard form of the circle so we have stopped the procedure when we got the circle equation as (x−h)2+(y−k)2=r2{{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}}. But when they have asked to find the equation of the circle then we need to expand the square terms by using the algebraic formulas either (a+b)2=a2+2ab+b2{{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}} or (a−b)2=a2−2ab+b2{{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}} and simplify the equation to get the equation of the circle.