Question

Find the centre and radius of the circle $\sqrt{1+{{m}^{2}}}\left( {{x}^{2}}+{{y}^{2}} \right)-2cx-2mcy=0$ .

Answer 1

Hint: We will be using the concepts of circle and coordinate geometry to solve the problem we will be using the general equation of circle with centre at $\left( h,k \right)$ and radius r is ${{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}}$.

Complete step-by-step solution -
 Now we have been given an equation of circle as $\sqrt{1+{{m}^{2}}}\left( {{x}^{2}}+{{y}^{2}} \right)-2cx-2mcy=0$.We have to find its centre and radius.
Now, we know that the general equation of a circle is ${{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}}$.
Where $\left( h,k \right)$is centre and r is radius. So, we will convert the given equation in this form and comparing we will get the centre and radius. $\sqrt{1+{{m}^{2}}}\left( {{x}^{2}}+{{y}^{2}} \right)-2cx-2mcy=0$.
Now, we will divide the whole equation by \[\sqrt{1+{{m}^{2}}}\] so that we have,
\[{{x}^{2}}+{{y}^{2}}-\dfrac{2c}{\sqrt{1+{{m}^{2}}}}x-\dfrac{2mc}{\sqrt{1+{{m}^{2}}}}y=0\]
Now, we will be completing the square with respect to x and y therefore,
\[{{x}^{2}}+{{y}^{2}}-\dfrac{2c}{\sqrt{1+{{m}^{2}}}}x-\dfrac{2mc}{\sqrt{1+{{m}^{2}}}}y=0\]
We have to take the half of the coefficients of x and y and then take its square. So, we will have $\dfrac{{{c}^{2}}}{1+{{m}^{2}}}$, $\dfrac{{{m}^{2}}{{c}^{2}}}{1+{{m}^{2}}}$. Now, we will add both of these on the L.H.S & R.H.S. Therefore, we have,
\[{{x}^{2}}-\dfrac{2c}{\sqrt{1+{{m}^{2}}}}x+{{\left( \dfrac{c}{\sqrt{1+{{m}^{2}}}} \right)}^{2}}+{{y}^{2}}-\dfrac{2mc}{\sqrt{1+{{m}^{2}}}}y+{{\left( \dfrac{mc}{\sqrt{1+{{m}^{2}}}} \right)}^{2}}=\dfrac{{{c}^{2}}}{1+{{m}^{2}}}+\dfrac{{{m}^{2}}{{c}^{2}}}{1+{{m}^{2}}}\]
Now, we will complete the square as we know that ${{a}^{2}}+{{b}^{2}}-2ab={{\left( a-b \right)}^{2}}$ . Therefore, we have,
\[{{\left( x-\dfrac{c}{\sqrt{1+{{m}^{2}}}} \right)}^{2}}+{{\left( y-\dfrac{mc}{\sqrt{1+{{m}^{2}}}} \right)}^{2}}=\dfrac{{{c}^{2}}\left( 1+{{m}^{2}} \right)}{1+{{m}^{2}}}\]
\[\Rightarrow {{\left( x-\dfrac{c}{\sqrt{1+{{m}^{2}}}} \right)}^{2}}+{{\left( y-\dfrac{mc}{\sqrt{1+{{m}^{2}}}} \right)}^{2}}={{c}^{2}}\]
Now on comparing with ${{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}}$
We have centre as \[\left( \dfrac{c}{\sqrt{1+{{m}^{2}}}},\dfrac{mc}{\sqrt{1+{{m}^{2}}}} \right)\] and radius as c.

Note: To solve these types of questions one should know the general equation of circle ${{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}}$. Also, it is important to note how we have completed the square with respect to x and y to obtain the equation in general forms. It should also be noted that the general equation of a circle is \[{{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0\] .