Question

The number of circle having radius 5 and passing through the point $\left( -2,0 \right)$ and $\left( 4,0 \right)$ is $A. one$
B. two $C. four$
D. infinite $$$$

Answer 1

We assume the centre as $\left( h,k \right)$. We use standard equation of circle in centre-radius form ${{\left( x-a \right)}^{2}}+{{\left( y-b \right)}^{2}}={{r}^{2}}$ and put $\left( -2,0 \right)$ and $\left( 4,0 \right)$ in the equation to find two equations in ${{h}^{2}},{{k}^{2}}$. We find the number of possible centres of $\left( h,k \right)$to find the number of circles. $$$$

Complete step by step answer:

Let us denote the centre be $\left( h,k \right)$.We are given the question that the radius is 5 units. We use standard equation of circle in centre-radius from to have the equation of the circle as

& {{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{5}^{2}} \\\ & \Rightarrow {{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}=25 \\\ \end{aligned}$$ We are also given the question that the above circle passes through the points $\left( -2,0 \right)$ and $\left( 4,0 \right)$. Since the point $\left( -2,0 \right)$ will satisfy equation of above circle, we have; $$\begin{aligned} & {{\left( -2-h \right)}^{2}}+{{\left( 0-k \right)}^{2}}=25 \\\ & \Rightarrow {{\left( h+2 \right)}^{2}}+{{k}^{2}}=25....\left( 1 \right) \\\ \end{aligned}$$ Since the point $\left( 4,0 \right)$ will satisfy equation of above circle, we have; $$\begin{aligned} & {{\left( 4-h \right)}^{2}}+{{\left( 0-k \right)}^{2}}=25 \\\ & \Rightarrow {{\left( 4-h \right)}^{2}}+{{k}^{2}}=25....\left( 2 \right) \\\ \end{aligned}$$ We equate the left hand sides of the equation (1) and (2) since right hand sides are equal to have; $$\begin{aligned} & {{\left( h+2 \right)}^{2}}+{{k}^{2}}={{\left( 4-h \right)}^{2}}+{{k}^{2}} \\\ & \Rightarrow {{\left( h+2 \right)}^{2}}-{{\left( 4-h \right)}^{2}}=0 \\\ \end{aligned}$$ We use the algebraic identity ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ for $a=h+2,b=4-h$ in the above step to have; $$\begin{aligned} & \Rightarrow \left( h+2-4+h \right)\left( h+2+4-h \right)=0 \\\ & \Rightarrow \left( 2h-2 \right)6=0 \\\ & \Rightarrow h=1 \\\ \end{aligned}$$ We put the values of $h=1$ in equation (1) to have; $$\begin{aligned} & {{\left( 1+2 \right)}^{2}}+{{k}^{2}}=25 \\\ & \Rightarrow {{k}^{2}}=25-{{3}^{2}}=16 \\\ & \Rightarrow k=4,-4 \\\ \end{aligned}$$ So we have two possible centres for the circle $\left( 1,4 \right),\left( 1,-4 \right)$. So there are two circles having radius 5 and passing through the point $\left( -2,0 \right)$ and $\left( 4,0 \right)$.  **So, the correct answer is “Option B”.** **Note:** We note that line segment joining the points $\left( -2,0 \right),\left( 4,0 \right)$whose length we can find using the distance formula as 6 units. Since the radius of the circle is given as 5 units, hence the chord is not a diameter. Now there are only two possibilities: either the centre is above the chord or below the chord, from where we get two circles.