Question

Find the equation of a circle passing through the origin and cutting the circle \[{{x}^{2}}+{{y}^{2}}+2{{g}_{1}}x+2{{f}_{1}}y+{{c}_{1}}=0\] and \[{{x}^{2}}+{{y}^{2}}+2{{g}_{2}}x+2{{f}_{2}}y+{{c}_{2}}=0\] orthogonally.

Answer 1

Hint: To solve this question, we should know the basic concept of the orthogonality condition which states that if two circles cuts each other orthogonally having equation of circle as \[{{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0\] and \[{{x}^{2}}+{{y}^{2}}+2g'x+2f'y+c'=0\] follows the relation of \[2g\times g'+2f\times f'=c+c'\].

Complete step-by-step answer:
We are given with two equation of circle let, C1 be \[{{x}^{2}}+{{y}^{2}}+2{{g}_{1}}x+2{{f}_{1}}y+{{c}_{1}}=0\] and C2 be \[{{x}^{2}}+{{y}^{2}}+2{{g}_{2}}x+2{{f}_{2}}y+{{c}_{2}}=0\].
Let us consider, the equation of circle which is cutting C1 and C2 orthogonally be C: \[{{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0\]. As we have given that the circle passes through origin then at (0,0), the equation of circle satisfies the condition, which implies, \[c=0\].
We know that when a circle, C cuts the other circle, C’ orthogonally then it follows the condition \[2g\times g'+2f\times f'=c+c'\].
Now, we first take circle C1, as C is cutting it orthogonally then according orthogonality criteria, we get,
\[2g\times {{g}_{1}}+2f\times {{f}_{1}}=c+{{c}_{1}}......\left( i \right)\]
Now, let us consider second circle C2, as C is cutting C2 orthogonally then according to orthogonality criteria, we get,
\[2g\times {{g}_{2}}+2f\times {{f}_{2}}=c+{{c}_{2}}......\left( ii \right)\]
Now, we will use elimination method to find the value of g and f because earlier we have found that \[c=0\].
After elimination, we found that \[f=\dfrac{{{c}_{1}}{{g}_{2}}-{{c}_{2}}{{g}_{1}}}{2\left( {{f}_{1}}{{g}_{2}}-{{f}_{2}}{{g}_{1}} \right)}\] and \[g=\dfrac{{{c}_{1}}{{f}_{2}}-{{c}_{2}}{{f}_{1}}}{2\left( {{f}_{1}}{{g}_{2}}-{{f}_{2}}{{g}_{1}} \right)}\].
Now, we are putting the values of ‘f’ and ‘g’ in the equation of circle, C.
Therefore, we get, \[{{x}^{2}}+{{y}^{2}}+2\left( \dfrac{{{c}_{1}}{{f}_{2}}-{{c}_{2}}{{f}_{1}}}{2\left( {{f}_{1}}{{g}_{2}}-{{f}_{2}}{{g}_{1}} \right)} \right)x+2\left( \dfrac{{{c}_{1}}{{g}_{2}}-{{c}_{2}}{{g}_{1}}}{2\left( {{f}_{1}}{{g}_{2}}-{{f}_{2}}{{g}_{1}} \right)} \right)y+c=0\]
On simplifying, we will get, \[\left( {{f}_{1}}{{g}_{2}}-{{f}_{2}}{{g}_{1}} \right){{x}^{2}}+\left( {{f}_{1}}{{g}_{2}}-{{f}_{2}}{{g}_{1}} \right){{y}^{2}}+\left( {{c}_{1}}{{f}_{2}}-{{c}_{2}}{{f}_{1}} \right)x+\left( {{c}_{1}}{{g}_{2}}-{{c}_{2}}{{g}_{1}} \right)y+c=0\] as the equation of circle which cuts the other circle orthogonally.

Note: We can also use a tangent method to find the equation of the circle but that will also come to the same point but will complicate the solution.