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# Find the equation of the circle with centre (0, 2) and radius 2 and center (-2, 3) and radius 4.

Last updated date: 22nd Jun 2024
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Hint: Now we know that the equation of a circle with center (h, k) and radius r is given by the ${{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}}$ . Now we know the center and radius of the circle hence by substituting the value of h, k and r we will get the equation of the required circle.

Now we are given that the center and radius of the circle.
Let us first understand the meaning of the equation of circle.
Equation of a circle is a general equation of all the points (x, y) on the circle.
Now a circle is formed by plotting all the points which are equidistant from one point which is center.
Now equation of circle with center as origin is given by ${{x}^{2}}+{{y}^{2}}={{r}^{2}}$ where r is the radius of the circle.
Now if the circle has center (h, k) is the center of the circle then the equation of circle is given by ${{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}}$ .
Now let us find the equation of the circle with center (0, 2) and radius 2.
Hence here we have h = 0, k = 2 and r = 2.
Hence the equation of the circle is,
$\Rightarrow {{x}^{2}}+{{\left( y-2 \right)}^{2}}={{2}^{2}}$
Now we know that ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$
\begin{align} & \Rightarrow {{x}^{2}}+{{y}^{2}}-2y+4=4 \\ & \Rightarrow {{x}^{2}}+{{y}^{2}}-2y=0 \\ \end{align}
Hence the equation of the circle is ${{x}^{2}}+{{y}^{2}}-2y=0$
Now consider the circle with center (-2, 3) and radius 4.
Here h = -2, k = 3 and r = 4.
Hence the equation of the circle is
$\Rightarrow {{\left( x-\left( -2 \right) \right)}^{2}}+{{\left( y-3 \right)}^{2}}={{\left( 4 \right)}^{2}}$
Now we know that ${{\left( a\pm b \right)}^{2}}={{a}^{2}}\pm 2ab+{{b}^{2}}$
\begin{align} & \Rightarrow {{x}^{2}}+2x+4+{{y}^{2}}-6y+9=16 \\ & \Rightarrow {{x}^{2}}+{{y}^{2}}+2x-6y-7=0 \\ \end{align}
Hence the equation of the circle is ${{x}^{2}}+{{y}^{2}}+2x-6y-7=0$ .

Note: Now note that the general equation of circle is ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$ where (-g, -f) is the center of the circle and ${{g}^{2}}+{{f}^{2}}-c$ is the radius of the circle. Hence the circle is real only if the value of ${{g}^{2}}+{{f}^{2}}>c$ . In other cases the circle is a point or imaginary circle.