Question

If a circle of radius r is touching the lines ${x^2} - 4xy + {y^2} = 0$ in the first quadrant at points A and B, then the area of triangle OAB (O being the origin) is:
$
  (1)\dfrac{{3\sqrt 3 {r^2}}}{4} \\
  (2)\dfrac{{\sqrt 3 {r^2}}}{4} \\
  (3)\dfrac{{3{r^2}}}{4} \\
  (4){r^2} \\
$

Answer 1

Hint: Find the individual lines from the combined equation of the pair of straight lines and then find the angle between the lines. Then draw a diagram to find angles and lengths to find the area.

Complete step-by-step answer:
To find the individual equations from ${x^2} - 4xy + {y^2} = 0$
Try to make square in one variable, by halving the coefficient of y and adding and subtracting the square of the halved term
$
  {y^2} - 4xy + 4{x^2} - 4{x^2} + {x^2} = 0 \\
  {(y - 2x)^2} = 3{x^2} \\
  y - 2x = \pm \sqrt 3 x \\
$
So the two lines are,
L1: $y = (2 + \sqrt 3 )x$ and L2: $y = (2 - \sqrt 3 )x$
${m_1} = 2 + \sqrt 3 {\text{ and }}{{\text{m}}_2} = 2 - \sqrt 3 $

Finding the angle between these lines using formula:
$\tan \theta = \dfrac{{|{m_2} - {m_1}|}}{{1 + {m_1}{m_2}}}$
 After substituting, we get $\tan \theta = \sqrt 3 {\text{ so }}\theta {\text{ = }}\dfrac{\pi }{3} = {60^ \circ }$

OS is the angle bisector of the internal angle between the lines OA and OB and it meets the centre of the circle.
We know that AS will be perpendicular to OA by the property that radius is always perpendicular to the tangent and the lines OA and OB are the tangents.
To find the area of $\Delta OAB$ , let's find the area of $\Delta OAM$ which we can find by taking the difference in the areas of $\Delta OAS{\text{ }}and{\text{ }}\Delta AMS$ .

Area of
 $
  \Delta OAS = \dfrac{1}{2} \times z \times r \\
  \Delta AMS = \dfrac{1}{2} \times x \times y \\
 $
Now, From
$
  \Delta OAS \\
  z = r\tan {60^ \circ } \\
  z = r\sqrt 3 \\
  \Delta AMS \\
  x = r\sin {60^ \circ } \\
  x = r\dfrac{{\sqrt 3 }}{2} \\
  y = r\cos {60^ \circ } \\
  y = \dfrac{r}{2} \\
 $
Area of $\Delta OAB = 2 \times (\Delta OAS - \Delta AMS)$
On putting the values in
  $
  \Delta OAS = \dfrac{1}{2} \times z \times r \\
  \Delta AMS = \dfrac{1}{2} \times x \times y \\
 $
We get
$
  2[ \times (\dfrac{1}{2} \times r \times \sqrt 3 r) - (\dfrac{1}{2} \times \dfrac{1}{2}r \times \dfrac{{\sqrt 3 }}{2}r)] \\
   = \dfrac{{3\sqrt 3 }}{4}{r^2} \\
$
Hence, option (1) is correct.

Note:
Deriving the individual equations from the combined equations should be known.
Properties of tangents are important in this case else the angles may go unnoticed and the question would become difficult to solve.