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

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Last updated date: 04th Mar 2024
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IVSAT 2024
Answer
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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.

Complete step by step answer:
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.
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