Question

Find the equation of the circle which is orthogonal to each of the following circles.
${{x}^{2}}+{{y}^{2}}+2x+4y+1=0$
$2\left( {{x}^{2}}+{{y}^{2}} \right)+6x+8y-3=0$
${{x}^{2}}+{{y}^{2}}-2x+6y-3=0$

Answer 1

Hint: To solve the above question, we will use the fact that if two circles ${{S}_{1}}=0$ and ${{S}_{2}}=0$ such that ${{S}_{1}}={{x}^{2}}+{{y}^{2}}+2{{g}_{1}}x+2{{f}_{1}}y+{{c}_{1}}$ and ${{S}_{2}}={{x}^{2}}+{{y}^{2}}+2{{g}_{2}}x+2{{f}_{2}}y+{{c}_{2}}$ then we can say that, $2{{g}_{1}}{{g}_{2}}+2{{f}_{1}}{{f}_{2}}={{c}_{1}}+{{c}_{2}}.$ With the help of this we will get three equations in three variables (f, g, x) which we will solve to get these variables. Then we will put these variables in the equation ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$ to get the required circle.

Complete step by step solution:
To start with, we will assume that the required equation of the circle is $S={{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0.$ Now, there are three circles which are orthogonal to these circle ‘S’. Before solving the question further, we must know what is orthogonal. Two circles are said to be orthogonal or said to intersect orthogonally if the tangents at their point of intersection include a right angle. If two circles are given such that,
$A={{x}^{2}}+{{y}^{2}}+2{{g}_{1}}x+2{{f}_{1}}y+{{c}_{1}}=0$
$B={{x}^{2}}+{{y}^{2}}+2{{g}_{2}}x+2{{f}_{2}}y+{{c}_{2}}=0$
Then they will be orthogonal only if the following condition is satisfied.
$2{{g}_{1}}{{g}_{2}}+2{{f}_{1}}{{f}_{2}}={{c}_{1}}+{{c}_{2}}$
The equations of the circles are given in the question are
${{S}_{1}}={{x}^{2}}+{{y}^{2}}+2x+4y+1=0......\left( i \right)$
${{S}_{2}}=2\left( {{x}^{2}}+{{y}^{2}} \right)+6x+8y-3=0$
${{S}_{2}}={{x}^{2}}+{{y}^{2}}+3x+4y-\dfrac{3}{2}=0.....\left( ii \right)$
${{S}_{3}}={{x}^{2}}+{{y}^{2}}-2x+6y-3=0.....\left( iii \right)$
Now S and ${{S}_{1}}$ are orthogonal, so,
$2g{{g}_{1}}+2f{{f}_{1}}=c+{{c}_{1}}$
In our case, $2{{g}_{1}}=2$ and $2{{f}_{1}}=4$ and ${{c}_{1}}=1.$ Thus,
$\Rightarrow 2g+4f=c+1$
$\Rightarrow 2g+4f-1=c.....\left( iv \right)$
Now, S and ${{S}_{2}}$ are orthogonal, so,
$2g{{g}_{2}}+2f{{f}_{2}}=c+{{c}_{2}}$
In our case, $2{{g}_{2}}=3,2{{f}_{2}}=4$ and ${{c}_{2}}=\dfrac{-3}{2}.$ Thus,
$\Rightarrow 3g+4f=c-\dfrac{3}{2}$
$\Rightarrow 3g+4f+\dfrac{3}{2}=c.....\left( v \right)$
Now, S and ${{S}_{3}}$ are orthogonal, so,
$2g{{g}_{3}}+2f{{f}_{3}}=c+{{c}_{3}}$
In our case, $2{{g}_{3}}=-2,2{{f}_{3}}=6$ and ${{c}_{3}}=-3.$ Thus,
$\Rightarrow -2g+6f=c-3$
$\Rightarrow -2g+6f+3=c.....\left( vi \right)$
From (iv) and (v), we have,
$\Rightarrow 2g+4f-1=c=3g+4f+\dfrac{3}{2}$
$\Rightarrow 2g+4f-1=3g+4f+\dfrac{3}{2}$
$\Rightarrow 2g-3g+4f-4f=1+\dfrac{3}{2}$
$\Rightarrow -g=\dfrac{5}{2}$
$\Rightarrow g=-\dfrac{5}{2}.....\left( vii \right)$
Now, we will put the value of g from (vii) to (vi). Thus, we will get,
$\Rightarrow -2\left( \dfrac{-5}{2} \right)+6f+3=c$
$\Rightarrow 5+6f+3=c$
$\Rightarrow c=6f+8.....\left( viii \right)$
Now, we will put the value of c from (viii) to (iv). Thus, we will get,
$2\left( \dfrac{-5}{2} \right)+4f-1=6f+8$
$\Rightarrow -5+4f-1=6f+8$
$\Rightarrow 4f-6=6f+8$
$\Rightarrow 4f-6f=6+8$
$\Rightarrow -2f=14$
$\Rightarrow f=-7.....\left( ix \right)$
Now, we will put the value of g and f from (vii) and (ix) to (iv). This, we will get,
$2\left( \dfrac{-5}{2} \right)+4\left( -7 \right)-1=c$
$\Rightarrow -5-28-1=c$
$\Rightarrow c=-34.....\left( x \right)$
Now, we will put the values of f, g and c in the equation of the circle S.
$S={{x}^{2}}+{{y}^{2}}+2\left( \dfrac{-5}{2} \right)x+2\left( -7 \right)y-34=0$
$S={{x}^{2}}+{{y}^{2}}-5x+-14y-34=0$
Thus, the equation of the circle which is orthogonal to the given three circles is
$S={{x}^{2}}+{{y}^{2}}-5x-14y-34=0$

Note: The alternate solution of this question is shown. The centre of the circle which is orthogonal to three given circles is the radical centre. For finding the centre of the equation, we will find the radical centre of ${{S}_{1}},{{S}_{2}}$ and ${{S}_{3}}.$ First we will subtract ${{S}_{1}}$ from ${{S}_{2}}.$ We will get,
${{S}_{2}}-{{S}_{1}}=\left( {{x}^{2}}+{{y}^{2}}+3x+4y-\dfrac{3}{2} \right)-\left( {{x}^{2}}+{{y}^{2}}+2x+4y+1 \right)=0$
$\Rightarrow 3x-2x-\dfrac{3}{2}-1=0$
$\Rightarrow x=\dfrac{5}{2}....\left( i \right)$
Now, we will subtract ${{S}_{1}}$ from ${{S}_{3}}.$ Thus, we will get,
${{S}_{3}}-{{S}_{1}}=\left( {{x}^{2}}+{{y}^{2}}-2x+6y-3 \right)-\left( {{x}^{2}}+{{y}^{2}}+2x+4y+1 \right)=0$
$\Rightarrow -4x+2y-4=0$
$\Rightarrow -2x+y-2=0$
$\Rightarrow y=2x+2$
$\Rightarrow y=2\left( \dfrac{5}{2} \right)+2\left[ \text{From (i)} \right]$
$\Rightarrow y=7$
Thus the centre of the equation is $\left( \dfrac{5}{2},7 \right).$ The equation of the circle can be given by,
$S={{\left( x-\dfrac{5}{2} \right)}^{2}}+{{\left( y-7 \right)}^{2}}={{r}^{2}}.....\left( ii \right)$
$\Rightarrow S={{x}^{2}}+{{y}^{2}}-5x+14y+\dfrac{25}{4}+49-{{r}^{2}}=0$
Now, we know that S and ${{S}_{1}}$ are orthogonal, so
$2{{g}_{1}}{{g}_{2}}+2{{f}_{1}}{{f}_{2}}={{c}_{1}}+{{c}_{2}}$
$\Rightarrow 2\times 1\times \left( \dfrac{-5}{2} \right)+2\times 2\times \left( -7 \right)=1+\dfrac{25}{4}+49-{{r}^{2}}$
$\Rightarrow -5-28=1+\dfrac{25}{4}+49-{{r}^{2}}$
$\Rightarrow {{r}^{2}}=\dfrac{357}{4}$
On putting the value of ${{r}^{2}}$ in (i), we get,
$S={{\left( x-\dfrac{5}{2} \right)}^{2}}+{{\left( y-7 \right)}^{2}}=\dfrac{357}{4}$
$\Rightarrow S={{x}^{2}}+{{y}^{2}}-5x-14y-34=0$