Question

A circle is described to pass through the origin and to touch the lines $ x = 1,x + y = 2. $ Prove that the radius of the circle is root of the equation $ \left( {3 - 2\sqrt 2 } \right){r^2} - 2\sqrt 2 r + 2 = 0 $ .

Answer 1

Hint: To solve this question, we use the general equation of the circle to find the standard formula for radius. We also know the lines touching the circles. Hence we can also calculate the radius by calculating the perpendicular distance of the line from the centre.

Complete step-by-step answer:
The general equation of the circle is
 $ {x^2} + {y^2} + 2gx + 2fy + c = 0 $
Where the centre point is $ ( - g, - f) $ and the formula to calculate the radius is $ \sqrt {{g^2} + {f^2}} $ .
Since the circle passes through the line $ x = 1 $ and $ x + y = 2 $ , we will calculate the perpendicular distance of these lines to the centre of the circle . This distance will be the value of radius( assume radius to be r).
Formula used:
 $ d = \dfrac{{A{x_1} + B{y_1} + C}}{{\sqrt {{A^2} + {B^2}} }} $
Perpendicular distance of $ ( - g, - f) $ from $ x = 1 $
 $
  r = \dfrac{{\left| { - g - 1} \right|}}{{\sqrt 1 }} \\
  r = g + 1 \;
  $
Perpendicular distance of $ ( - g, - f) $ from $ x + y = 2 $
 $
  r = \dfrac{{\left| { - g - f - 2} \right|}}{{\sqrt 2 }} \\
  \sqrt 2 r = g + f + 2 \;
  $
Comparing the above two equations
 $
  \sqrt 2 r = g + f + 2 \\
  \sqrt 2 r = r - 1 + f + 2 \\
  \sqrt 2 r = r + f + 1 \\
  f = (\sqrt 2 - 1)r - 1 \;
  $
Now , we know that
 $
  {r^2} = {f^2} + {g^2} \\
  {r^2} = {((\sqrt 2 - 1)r - 1)^2} + {(r - 1)^2} \\
  {r^2} = {(\sqrt 2 - 1)^2}{r^2} + 1 - 2(\sqrt 2 - 1)r + {r^2} + 1 - 2r \\
  0 = {(\sqrt 2 - 1)^2}{r^2} + 2 - 2\sqrt 2 r \\
  0 = (2 + 1 - 2\sqrt 2 ){r^2} + 2 - 2\sqrt 2 r \\
  0 = (3 - 2\sqrt 2 ){r^2} + 2 - 2\sqrt 2 r \;
  $
Hence, the radius of the circle is the root of the following equation $ 0 = (3 - 2\sqrt 2 ){r^2} + 2 - 2\sqrt 2 r $

Note: If the value of $ {f^2} + {g^2} $ is greater than c , then the radius of the circle is real. If the value of $ {f^2} + {g^2} $ is less than c , then the radius of the circle is imaginary. And if the value of $ {f^2} + {g^2} $ is equal to zero, that means the radius is zero. Hence there is no circle but it is a point.