Question

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.

Answer 1

Hint: 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 \[{x^2} + {y^2} + 2{\text{g}}x + 2{\text{f}}y + c = 0\]
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
\[
  {a^2} + {b^2} + 2{\text{ga}} + 2{\text{gb}} + c = 0 \\
   \Rightarrow {a^2} + {b^2} + 2{\text{g}}\left( {{\text{a + b}}} \right) = 0 \\
   \Rightarrow {\text{g = - }}\dfrac{{\left( {{a^2} + {b^2}} \right)}}{{2\left( {{\text{a + b}}} \right)}} \\
 \]\[{a^2} + {b^2} + 2{\text{ga}} + 2{\text{fb}} + c = 0 - - - \left( 1 \right)\]
Now substituting the value of (b, a) in the general equation, we get
\[{b^2} + {a^2} + 2{\text{gb}} + 2{\text{fa}} + c = 0 - - - \left( 2 \right)\]
Lastly substituting the value (0, 0) in general equation, we get
\[
  {0^2} + {0^2} + 2{\text{g}} \times {\text{0}} + 2{\text{f}} \times {\text{0}} + c = 0 \\
   \Rightarrow c = 0 - - - \left( 3 \right) \\
 \]
Substituting the value of c=0 in equation (1) and (2) and solving equation (1) and (2) we get\[
  {a^2} + {b^2} + 2{\text{g}}a + 2{\text{f}}b + c = {b^2} + {a^2} + 2{\text{g}}b + 2{\text{f}}a + c = 0 \\
   \Rightarrow 2{\text{g}}a + 2{\text{f}}b = 2{\text{g}}b + 2{\text{f}}a \\
   \Rightarrow {\text{g}}a + {\text{f}}b = {\text{g}}b + {\text{f}}a \\
   \Rightarrow {\text{g}}a - {\text{g}}b = {\text{f}}a - {\text{f}}b \\
   \Rightarrow {\text{g}} = {\text{f}} \\
 \]
Substituting this value again in equation (1) we get
\[
  {a^2} + {b^2} + 2{\text{g}}a + 2{\text{g}}b + c = 0 \\
   \Rightarrow {a^2} + {b^2} + 2{\text{g}}\left( {a + b} \right) = 0 \\
   \Rightarrow {\text{g}} = - \dfrac{{\left( {{a^2} + {b^2}} \right)}}{{2\left( {a + b} \right)}} \\
 \]
$ \Rightarrow {\text{g = f = }}\dfrac{{ - \left( {{a^2} + {b^2}} \right)}}{{2\left( {a + b} \right)}}$
So substituting the above values in general equation we get\[
   \Rightarrow {x^2} + {y^2} + 2\left( {\dfrac{{ - \left( {{a^2} + {b^2}} \right)}}{{2\left( {a + b} \right)}}} \right)x + 2\left( {\dfrac{{ - \left( {{a^2} + {b^2}} \right)}}{{2\left( {a + b} \right)}}} \right)y = 0 - - - (4) \\
   \Rightarrow {x^2} + {y^2} - \dfrac{{\left( {{a^2} + {b^2}} \right)}}{{\left( {a + b} \right)}}\left( {x + y} \right) = 0 \\
 \]
Therefore the equation of the circle is \[{x^2} + {y^2} - \dfrac{{\left( {{a^2} + {b^2}} \right)}}{{\left( {a + b} \right)}}\left( {x + y} \right) = 0\]
Now the general equation to find the value of x-intercept and y-intercept is given by
$
  x - {\text{intercept = 2}}\sqrt {{{\text{g}}^{\text{2}}}{\text{ - c}}} \\
  y - {\text{intercept = 2}}\sqrt {{{\text{f}}^{\text{2}}}{\text{ - c}}} \\
 $
So substituting the values of g, f and c respectively we get
$
  x - {\text{intercept = 2}}\sqrt {{{\text{g}}^{\text{2}}}{\text{ - c}}} \\
   = 2\sqrt {{{\left( {\dfrac{{ - \left( {{a^2} + {b^2}} \right)}}{{2\left( {a + b} \right)}}} \right)}^2}} \\
   = \dfrac{{\left( {{a^2} + {b^2}} \right)}}{{\left( {a + b} \right)}} \\
  y - {\text{intercept = 2}}\sqrt {{{\text{f}}^{\text{2}}}{\text{ - c}}} \\
   = 2\sqrt {{{\left( {\dfrac{{ - \left( {{a^2} + {b^2}} \right)}}{{2\left( {a + b} \right)}}} \right)}^2}} \\
   = \dfrac{{\left( {{a^2} + {b^2}} \right)}}{{\left( {a + b} \right)}} \\
 $
Therefore the values of x-intercept and y-intercept are $\dfrac{{\left( {{a^2} + {b^2}} \right)}}{{\left( {a + b} \right)}}$ respectively.

Note: This kind of solution can be easily solved with the help of the general form of the equation of circle. Always try to start from the information given to us rather than focusing on what we have to find. This makes the process a little easier. Also try to write the final equation of the circle similar to the general equation.