Question
Question: Find the equation to the circle passing through the origin and points (a, b) and (b, a). Find the le...
Find the equation to the circle passing through the origin and points (a, b) and (b, a). Find the length of the chords that it cuts off the axes.
Solution
Given in the question is the point the circle is passing through. To solve such a question, try to list down the general formula of the circle and substitute in all the values given individually. After getting the equations try to solve the equations to get the equation of the circle and the x-intercept and y-intercept respectively.
Complete step-by-step solution:
We will start the solution by noting down the general form of circle
The general form of circle is given by x2+y2+2gx+2fy+c=0
It’s been given to us that the circle passes through the origin and the points (a, b) and (b, a).
That means these given points satisfy the general equation of the circle listed above for the points (a, b), (b, a) and (0, 0).
Therefore substituting the values of (a, b) in our general equation of circle we get
{0^2} + {0^2} + 2{\text{g}} \times {\text{0}} + 2{\text{f}} \times {\text{0}} + c = 0 \\
\Rightarrow c = 0 - - - \left( 3 \right) \\
{a^2} + {b^2} + 2{\text{g}}a + 2{\text{f}}b + c = {b^2} + {a^2} + 2{\text{g}}b + 2{\text{f}}a + c = 0 \\
\Rightarrow 2{\text{g}}a + 2{\text{f}}b = 2{\text{g}}b + 2{\text{f}}a \\
\Rightarrow {\text{g}}a + {\text{f}}b = {\text{g}}b + {\text{f}}a \\
\Rightarrow {\text{g}}a - {\text{g}}b = {\text{f}}a - {\text{f}}b \\
\Rightarrow {\text{g}} = {\text{f}} \\
{a^2} + {b^2} + 2{\text{g}}a + 2{\text{g}}b + c = 0 \\
\Rightarrow {a^2} + {b^2} + 2{\text{g}}\left( {a + b} \right) = 0 \\
\Rightarrow {\text{g}} = - \dfrac{{\left( {{a^2} + {b^2}} \right)}}{{2\left( {a + b} \right)}} \\
\Rightarrow {x^2} + {y^2} + 2\left( {\dfrac{{ - \left( {{a^2} + {b^2}} \right)}}{{2\left( {a + b} \right)}}} \right)x + 2\left( {\dfrac{{ - \left( {{a^2} + {b^2}} \right)}}{{2\left( {a + b} \right)}}} \right)y = 0 - - - (4) \\
\Rightarrow {x^2} + {y^2} - \dfrac{{\left( {{a^2} + {b^2}} \right)}}{{\left( {a + b} \right)}}\left( {x + y} \right) = 0 \\