Question

I

if the parametric values of two points a and b lying on the circle x square + y square - 6x + 4y -12=0 are 30 degree and 60 degree respectively then find the equation of the chord joining A and B

Answer 1

Given :

Equation of the given circle is :

$x^2 + y^2 - 6x +4y -12 =0$

Parametric value of 1st point A on this circle = 30°

Parametric value of 2nd point B on this circle = 60°

To Find :

Equation of the  chord joining points A and B = ?

Solution :

From the given equation the centre of circle O is :

( 3 , -2 )

Radius of this circle is :

= $\sqrt {3^2 + (-2)^2 - (-12)}$

= $\sqrt{9 + 4 + 12 }$

=$\sqrt{25}$

= 5 units

Now drop perpendiculars from the points A and B on O .

Horizontal component of OA = 5 cos30 = $\frac{5\sqrt3}{2}$

Vertical component of OA = 5 sin30 = $\frac{5}{2}$

Similarly, horizontal comp. of OB = 5 cos60 =$\frac{5}{2}$

And vertical comp. of OB  = 5 sin60 = $\frac{5\sqrt3}{2}$

So, coordinates of A = ( $3 + \frac{5\sqrt3}{2} , -2 + \frac{5}{2}$ ) = ( $\frac{6 + 5\sqrt3}{2} ,\frac{1}{2}$ )

And coordinates of B = ( $3 + \frac{5}{2} , -2 + \frac{5\sqrt3}{2}$ ) = ( $\frac{11}{2} , \frac{5\sqrt3 - 4}{2}$ )

So, the equation of line joining A and B will be :

$\frac{x - \frac{11}{2}}{y - (\frac{5\sqrt3-4}{2})} = \frac{(\frac{5\sqrt3-4}{2}-\frac{1}{2}}{\frac{11}{2} - (\frac{6 + 5\sqrt3}{2})}$

Or, $\frac{2x-11}{2y - 5\sqrt3 +4}= \frac{5(\sqrt3 -1)}{5(1-\sqrt3)}$

Or, $2x -11 = -1 (2y - 5 \sqrt3 + 4)$

Or, $2x + 2y = 7 + 5 \sqrt3$

Hence , equation of the chord joining A and B is $2x + 2y = 7 + 5 \sqrt3$