Question

Prove that the circle through the origin and cutting circles
${x^2} + {y^2} + 2{g_1}x + 2{f_1}y + {c_1} = 0$ and ${x^2} + {y^2} + 2{g_2}x + 2{f_2}y + {c_2} = 0$
Orthogonally is \left| {\begin{array}{*{20}{c}} {{x^2} + {y^2}}&{ - x}&{ - y} \\\ {{c_1}}&{{g_1}}&{{f_1}} \\\ {{c_2}}&{{g_2}}&{{f_2}} \end{array}} \right| = 0

Answer 1

In this question remember to use formula of determinant i.e. $\left| \begin{gathered} a{\text{ }}b{\text{ }}c \\\ e{\text{ }}f{\text{ }}g \\\ h{\text{ }}i{\text{ }}j \\\ \end{gathered} \right| = a\left( {fj - gi} \right) - b\left( {ej - gh} \right) + c\left( {ei - fh} \right)$to find the equation of circle and remember that when one circle cuts the other circle orthogonally then $2{g_1}{g_2} + 2{f_1}{f_2} = {C_1} + {C_2}$, using this information will help you to approach the solution of the question.

Complete step-by-step answer:
Let’s first find out what is the equation of the circle we have in this determinant form
Solving the determinant: -
\left[ {({x^2} + {y^2})\left| {\begin{array}{*{20}{c}} {{g_1}}&{{f_1}} \\\ {{g_2}}&{{f_2}} \end{array}} \right|} \right] - \left[ {( - x)\left| {\begin{array}{*{20}{c}} {{c_1}}&{{f_1}} \\\ {{c_2}}&{{f_2}} \end{array}} \right|} \right] + \left[ {( - y)\left| {\begin{array}{*{20}{c}} {{c_1}}&{{g_1}} \\\ {{c_2}}&{{g_2}} \end{array}} \right|} \right] = 0
$\Rightarrow$ $({x^2} + {y^2})({g_1}{f_2} - {g_2}{f_1}) + x({c_1}{f_2} - {c_2}{f_1}) - y({c_1}{g_2} - {c_2}{g_1}) = 0$ (equation i)
This is the equation of the circle.
Since the circle is passing through the origin the equation will be
${x^2} + {y^2} + 2gx + 2fy + C = 0$
Since the circle is passing through origin, C = 0
Hence the equation of the circle will become
${x^2} + {y^2} + 2gx + 2fy = 0$
Now the circle is cutting the
${x^2} + {y^2} + 2{g_1}x + 2{f_1}y + {c_1} = 0$and ${x^2} + {y^2} + 2{g_2}x + 2{f_2}y + {c_2} = 0$ circle orthogonally
When one circle cuts the other circle orthogonally then $2{g_1}{g_2} + 2{f_1}{f_2} = {C_1} + {C_2}$
For circle with equation
${x^2} + {y^2} + 2gx + 2fy = 0$and ${x^2} + {y^2} + 2{g_1}x + 2{f_1}y + {c_1} = 0$
$\Rightarrow$ $2g{g_1} + 2f{f_1} = 0 + {C_1}$
$\Rightarrow$ $2g{g_1} + 2f{f_1} = {C_1}$(equation ii)
And for circle with equation
${x^2} + {y^2} + 2gx + 2fy = 0$and ${x^2} + {y^2} + 2{g_2}x + 2{f_2}y + {c_2} = 0$
$\Rightarrow$ $2g{g_2} + 2f{f_2} = 0 + {C_2}$
$\Rightarrow$ $2g{g_2} + 2f{f_2} = {C_2}$ (equation iii)
Now we have to solve for g and f from equation
$2g{g_1} + 2f{f_1} = {C_1}$and $2g{g_2} + 2f{f_2} = {C_2}$
For finding out the value of f we have to cancel out the value of g for this
We will multiply equation (ii) with ${g_2}$ and (iii) with ${g_1}$ and then subtract equation (iii) with equation (ii).
$(2g{g_1}{g_2} + 2f{f_2}{g_2} = {C_1}{g_2}) - (2g{g_1}{g_2} + 2f{f_2}{g_1} = {C_2}{g_1})$
$\Rightarrow$ $2f{f_1}{g_2} - 2f{f_2}{g_1} = {C_1}{g_2} - {C_2}{g_1}$Taking 2f common
$\Rightarrow$ $2f({f_1}{g_2} - {f_2}{g_1}) = {C_1}{g_2} - {C_2}{g_1}$
$\Rightarrow$ $2f = \dfrac{{({C_1}{g_2} - {C_2}{g_1})}}{{({f_1}{g_2} - {f_2}{g_1})}}$ (equation iv)
Now, for finding out the value g we have to cancel out the value of for this f
We will multiply equation (ii) with f2 and (iii) with f1 and then subtract equation (iii) with equation (ii).

$(2g{g_1}{f_2} + 2f{f_1}{f_2} = {C_1}{f_2}) - (2g{g_1}{f_1} + 2f{f_2}{f_1} = {C_2}{f_1})$
$2g{g_1}{f_2} - 2g{g_2}{f_1} = {C_1}{f_2} - {C_2}{f_1}$ Taking 2g common
$\Rightarrow$ $2g({g_1}{f_2} - {g_2}{f_1}) = {C_1}{f_2} - {C_2}{f_1}$
$\Rightarrow$ $2g = \dfrac{{({C_1}{f_2} - {C_2}{f_1})}}{{({g_1}{f_2} - {g_2}{f_1})}}$ (equation v)
By solving the equations, we get the value of g and f
Now the equation of circle becomes
${x^2} + {y^2} + 2gx + 2fy = 0$
Putting the value of g and f from (iv) and (v)
$({x^2} + {y^2}) + \left( {\dfrac{{({c_1}{f_2} - {c_2}{f_1})}}{{({g_1}{f_2} - {g_2}{f_1})}}} \right)x + \left( {\dfrac{{({c_1}{g_2} - {c_2}{g_1})}}{{({f_1}{g_2} - {f_2}{g_1})}}} \right)y = 0$
Multiplying each side by $({g_1}{f_2} - {g_2}{f_1})$
We get the equation
${x^2}({g_1}{f_2} - {g_2}{f_1}) + {y^2}({g_1}{f_2} - {g_2}{f_1}) + x({c_1}{f_2} - {c_2}{f_1}) - y({c_1}{g_2} - {c_2}{g_1}) = 0$
$({x^2} + {y^2})({g_1}{f_2} - {g_2}{f_1}) + x({c_1}{f_2} - {c_2}{f_1}) - y({c_1}{g_2} - {c_2}{g_1}) = 0$ (equation vi)
If we compare the equation (i) and (vi) then we will find that both are same
This indicated that \left| {\begin{array}{*{20}{c}} {{x^2} + {y^2}}&{ - x}&{ - y} \\\ {{c_1}}&{{g_1}}&{{f_1}} \\\ {{c_2}}&{{g_2}}&{{f_2}} \end{array}} \right| = 0 is the equation of the circle which is passing through the through the origin and cutting circles ${x^2} + {y^2} + 2{g_1}x + 2{f_1}y + {c_1} = 0$ and ${x^2} + {y^2} + 2{g_2}x + 2{f_2}y + {c_2} = 0$ Orthogonally

Note: In such types of problems it is very important to consider the understanding of the equation of circle. Equation of a circle is ${x^2} + {y^2} + 2gx + 2fy + C = 0$ and also when a circle pass through origin C = 0, we must remember the properties of the circle and the process to solve the determinant.