Question
The lines $$2x-3y = 5$$ and $$3x-4y = 7$$ are diameters of a circle of area $$154 \,sq.\, units.$$ Then the equation of this circle is-
A.
$${x^2} + {y^2} + 2x - 2y = 62$$
B.
$${x^2} + {y^2} + 2x - 2y = 47$$
C.
$${x^2} + {y^2} - 2x + 2y = 47$$
D.
$${x^2} + {y^2} - 2x + 2y = 62$$
Answer :
$${x^2} + {y^2} - 2x + 2y = 47$$
Solution :
As $$2x-3y-5=0$$ and $$3x-4y-7=0$$ are diameters of circles.
$$\therefore $$ Centre of circle is solution of these two equation $$'s,$$ i.e.
$$\eqalign{
& \frac{x}{{21 - 20}} = \frac{y}{{ - 15 + 14}} = \frac{1}{{ - 8 + 9}} \cr
& \Rightarrow x = 1,\,\,y = - 1 \cr
& \therefore C\left( {1,\, - 1} \right) \cr} $$
Also area of circle, $$\pi {r^2} = 154$$
$$\eqalign{
& \Rightarrow {r^2} = \frac{{154}}{{22}} \times 7 = 49 \cr
& \Rightarrow r = 7 \cr} $$
$$\therefore $$ Equation of required circle is
$$\eqalign{
& {\left( {x - 1} \right)^2} + {\left( {y + 1} \right)^2} = {7^2} \cr
& \Rightarrow {x^2} + {y^2} - 2x + 2y = 47 \cr} $$