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

Find the centre and radius of the circle 1+m2(x2+y2)−2cx−2mcy=0\sqrt{1+{{m}^{2}}}\left( {{x}^{2}}+{{y}^{2}} \right)-2cx-2mcy=0 .

Explanation

Solution

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 (h,k)\left( h,k \right) and radius r is (x−h)2+(y−k)2=r2{{\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 1+m2(x2+y2)−2cx−2mcy=0\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 (x−h)2+(y−k)2=r2{{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}}.
Where (h,k)\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. 1+m2(x2+y2)−2cx−2mcy=0\sqrt{1+{{m}^{2}}}\left( {{x}^{2}}+{{y}^{2}} \right)-2cx-2mcy=0.
Now, we will divide the whole equation by 1+m2\sqrt{1+{{m}^{2}}} so that we have,
x2+y2−2c1+m2x−2mc1+m2y=0{{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,
x2+y2−2c1+m2x−2mc1+m2y=0{{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 c21+m2\dfrac{{{c}^{2}}}{1+{{m}^{2}}}, m2c21+m2\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,
x2−2c1+m2x+(c1+m2)2+y2−2mc1+m2y+(mc1+m2)2=c21+m2+m2c21+m2{{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 a2+b2−2ab=(a−b)2{{a}^{2}}+{{b}^{2}}-2ab={{\left( a-b \right)}^{2}} . Therefore, we have,
(x−c1+m2)2+(y−mc1+m2)2=c2(1+m2)1+m2{{\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}}}
⇒(x−c1+m2)2+(y−mc1+m2)2=c2\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 (x−h)2+(y−k)2=r2{{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}}
We have centre as (c1+m2,mc1+m2)\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 (x−h)2+(y−k)2=r2{{\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 x2+y2+2gx+2fy+c=0{{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0 .