Answer

Verified

342.3k+ views

**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.

Recently Updated Pages

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

How do you arrange NH4 + BF3 H2O C2H2 in increasing class 11 chemistry CBSE

Is H mCT and q mCT the same thing If so which is more class 11 chemistry CBSE

What are the possible quantum number for the last outermost class 11 chemistry CBSE

Is C2 paramagnetic or diamagnetic class 11 chemistry CBSE

What happens when entropy reaches maximum class 11 chemistry JEE_Main

Trending doubts

How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

What is the color of ferrous sulphate crystals? How does this color change after heating? Name the products formed on strongly heating ferrous sulphate crystals. What type of chemical reaction occurs in this type of change.

Write the difference between soap and detergent class 10 chemistry CBSE

Give 10 examples of unisexual and bisexual flowers

Differentiate between calcination and roasting class 11 chemistry CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

What is the difference between anaerobic aerobic respiration class 10 biology CBSE

a Why did Mendel choose pea plants for his experiments class 10 biology CBSE