Question

(2) If the lines 2x – 3y = 5 and 3x – 4y= 7 are the
diameters of a circle of area 154 sq. units,
then find the equation of the circle.
(A)x2 + y2 – 2x + 2y = 40
(B) x2 + y2 – 2x – 2y = 47
(C) x2 + y2 – 2x + 2y = 47
(D) x2 + y2 – 2x – 2y = 40​

Answer 1

intersection of diameter is the point(1,-1)

πs²=154

s²=49

(x-1)²+(y+1)²=49

hence(C)is the correct option

Answer 2

Given

equation of diameter of line

2x – 3y = 5-(1)*3

3x – 4y= 7-(2)*2

area of circle= 154unit^2

we know that

area of circle = πr^2

according to question

22r^2/7 = 154

r^2= 14*7/2

r = 7

Diameter = 14

Since both line are diameter of circle so it satisfied the line

On subtracting above equation we get

y= -1

So

X = 1

We know that

equation of circle is

(x - a)^2 + (y - b)^2 = r^2

where

a = 1

b=1

r =7

Now

(X - 1)^2+ (y + 1)^2 = 49

x^2 + y^2 - 2x + 2y = 47

Option C