Question

If the circle $$x^2+y^2+2gx+2fy+c=0$$ cuts the three circles $$x^2+y^2-5=0,x^2+y^2-8x-6y+10=0$$ and $$x^2+y^2-4x+2y-2=0$$ at the extremities of their diameter, then which of the following is/are true?
(a) $$c=-5$$
(b) $$f.g={\displaystyle \frac{147}{25}}$$
(c) $$g+2f=c+2$$
(d) $$4f=3g$$

Answer 1

Hint: In order to solve this problem we need to use the concept of the orthogonal circle, orthogonal circles are those circles which intersect at right angles. We need to compare the coefficients of given circles with the general equation of the circle then use the condition of orthogonal of circles. The formula for condition of orthogonality:
$$\Rightarrow 2(g_1{g}_2+f_1{f}_2)=C_1+C_2$$

Complete step-by-step answer:
Orthogonal circles:
When the first circle intersects the three circles at the extremities of the diameter, then they are orthogonal to each other.
Given that the equations of three circles are
$$\Rightarrow C_1\equiv x^2+y^2-5=0,C_2\equiv x^2+y^2-8x-6y+10=0$$ and $$C_3\equiv x^2+y^2-4x-2y-2=0$$
The center of the first circle $$C_1$$, is at $$\left(0,0\right)$$.
For center of second circle, $$C_2$$, we have:
$$\begin{gathered} \Rightarrow 2g=-8 \\ \Rightarrow g={\displaystyle \frac{-8}{2}}=-4 \end{gathered}$$
and
$$\begin{gathered} \Rightarrow 2f=-6 \\ \Rightarrow f=-{\displaystyle \frac{6}{2}}=-3 \end{gathered}$$
Centre of the second circle, $$C_2$$, is at $$(-g,-f)=(4,3)$$.
Similarly for the center of third circle, $$C_3$$, we have:
$$\begin{gathered} \Rightarrow 2g=-4 \\ \Rightarrow g={\displaystyle \frac{-4}{2}}=-2 \end{gathered}$$
and
$$\begin{gathered} \Rightarrow 2f=2 \\ \Rightarrow f={\displaystyle \frac{2}{2}}=1 \end{gathered}$$
Centre of the second circle, $$C_3$$, is at $$(-g,-f)=(2,-1)$$.
The circle which cuts these circles is given by:
$$\Rightarrow x^2+y^2+2gx+2fy+c=0$$
Then its center is:
$$\Rightarrow C_4\equiv (-g,-f)$$
Constant term of circle $$C_1$$ is $$-5$$
Constant term of circle $$C_2$$ is $$10$$
Constant term of circle $$C_3$$ is $$-2$$
Constant term of the circle which cuts these circle i.e. $$C_4$$ is $$c$$
By using the concept of the orthogonal circle, we’ll get:
For the circle $$C_{1}and{C}_4$$
$$\begin{gathered} \Rightarrow 2(g_1{g}_4+f_1{f}_4)=c_1+c_4 \\ \Rightarrow 2(-g(0)+-f(0))=c-5 \\ \Rightarrow 0=c-5 \\ \Rightarrow c=5 \end{gathered}$$
Thus, option (a) is incorrect.
Similarly using the concept of the orthogonal for the circle $$C_{2}and{C}_4$$, we’ll get
$$\Rightarrow 2[g_2{g}_4+f_2{f}_4]=c_2+c_4$$
Rewrite the equation after simplification
$$\Rightarrow 2[4(-g)+3(-f)]=10+c......(1)$$
Substitute the value of c in equation $$(1)$$, we’ll get:
$$\begin{gathered} \Rightarrow 2[-4g-3f]=10+5 \\ \Rightarrow -4g-3f={\displaystyle \frac{15}{2}} \\ \Rightarrow 4g+3f=-{\displaystyle \frac{15}{2}}.....(2) \end{gathered}$$
Use the concept of the orthogonal circle for the circle $$C_{3}and{C}_4$$, we’ll get:
$$\begin{gathered} \Rightarrow 2[g_3{g}_4+f_3{f}_4]=c_3+c_4 \\ \Rightarrow 2[2(-g)+(-1)(-f)]=-2+c \end{gathered}$$
Rewrite the equation after simplification:
$$\begin{gathered} \Rightarrow -2g+f={\displaystyle \frac{-2+5}{2}} \\ \Rightarrow -2g+f={\displaystyle \frac{3}{2}} \end{gathered}$$
Simplifying the equation, we have:
$$\Rightarrow f={\displaystyle \frac{3}{2}}+2g....(3)$$
Substitute the value of the f in the equation $$(2)$$:
$$\begin{gathered} \Rightarrow 4g+3\left({\displaystyle \frac{3}{2}}+2g\right)=-{\displaystyle \frac{15}{2}} \\ \Rightarrow 4g+{\displaystyle \frac{9}{2}}+6g=-{\displaystyle \frac{15}{2}} \end{gathered}$$
Simplifying the equation we’ll get:
$$\begin{gathered} \Rightarrow 10g=-{\displaystyle \frac{15}{2}}-{\displaystyle \frac{9}{2}} \\ \Rightarrow 10g=-{\displaystyle \frac{24}{2}} \\ \Rightarrow g=-{\displaystyle \frac{12}{10}} \end{gathered}$$
Substitute the value of g back in equation $$(3)$$:
$$\begin{gathered} \Rightarrow f={\displaystyle \frac{3}{2}}+2\times \left(-{\displaystyle \frac{12}{10}}\right) \\ \Rightarrow f={\displaystyle \frac{3}{2}}-{\displaystyle \frac{24}{10}} \end{gathered}$$
Simplifying it further, we’ll get:
$$\begin{gathered} \Rightarrow f={\displaystyle \frac{15-24}{10}} \\ \Rightarrow f={\displaystyle \frac{-9}{10}} \end{gathered}$$
Therefore, the product of f and g is:
$$\begin{gathered} \Rightarrow f\cdot g=-{\displaystyle \frac{9}{10}}\times -{\displaystyle \frac{12}{10}} \\ \Rightarrow f\cdot g={\displaystyle \frac{9\times 6}{10\times 5}} \\ \Rightarrow f\cdot g={\displaystyle \frac{27}{25}} \end{gathered}$$
Therefore option (b) is incorrect.
Comparing the values of $$4f$$ and $$3g$$, we’ll get:
$$\begin{gathered} \Rightarrow 4f=4\times \left(-{\displaystyle \frac{9}{10}}\right) \\ \Rightarrow 4f=-{\displaystyle \frac{36}{10}} \end{gathered}$$
and $$3g=3\times \left(-{\displaystyle \frac{12}{10}}\right)=-{\displaystyle \frac{36}{10}}$$
So we get:
$$\Rightarrow 4f=3g$$
Hence, option (d) is the correct.
Now, comparing the value of $$g+2f$$ and $$c+2$$, we have:
$$\begin{gathered} \Rightarrow g+2f=-{\displaystyle \frac{12}{10}}-2\times {\displaystyle \frac{9}{10}} \\ \Rightarrow g+2f=-{\displaystyle \frac{30}{10}}=-3 \end{gathered}$$
Similarly we have:
$$\Rightarrow c+2=5+2=7$$
So we get:
$$\Rightarrow g+2f\ne c+2$$
Thus, option (c) is also incorrect.

Hence, only option (d) is correct.

Note: The general equation of circle is given as:
$$\Rightarrow x^2+y^2+2gx+2fy+c=0$$
From this general equation, we get the center of the circle as $$C\equiv (-g,-f)$$ and its radius is $$r=\sqrt{g^2+f^2-c}$$.
For a two degree equation in $$x$$ and $$y$$ to be circle, the coefficients of $$x^2$$ and $$y^2$$ must be the same.