The lines `2x-3y=5` and `3x-4y=7` are the diameters of a circle of area 154 sq. units. Then the equation of the circle is `x^2+y^2+2x-2y=62` `x^2+y^2+2x-2y=47` `x^2+y^2-2x+2y=47` `x^2+y^2-2x+2y=62`

If the lines 2x-3y=5 and 3x-4y=7 are the diameters of a circle of area 154 square units, then obtain the equation of the circle.

The lines 2x-3y=5 and 3x-4y=7 are diameters of a circle of are 154 sq. units. Taking pi=22/7 , show that the equation of the circle is x^(2)+y^(2)-2x+2y=47

If the lines 2x + 3y + 1 = 0 and 3x - y-4 = 0 lie along two diameters of a circle of circumference 10 pi , then the equation of circle is (i) x^(2) + y^(2) + 2x + 2y + 23 = 0 (ii) x^(2) + y^(2) - 2x - 2y - 23 = 0 (iii) x^(2) + y^(2) - 2 x + 2y - 23 = 0 (iv) x^(2) + y^(2) + 2x - 2y + 23 = 0

The radical centre of the circles x^2+y^2-x+3y-3=0 , x^2+y^2-2x+2y-2=0, x^2+y^2+2x+3y-9=0

Find the parametric equations of that circles : 2x^2 + 2y^2 - 5x - 7y-3=0

The radical centre of the circles x^2+y^2=9, x^2+y^2-2x-2y-5=0 , x^2+y^2+4x+6y-19=0 is

The point of tangency of the circles x^2+ y^2 - 2x-4y = 0 and x^2 + y^2-8y -4 = 0 , is

The point of tangency of the circles x^2+ y^2 - 2x-4y = 0 and x^2 + y^2-8y -4 = 0 , is

The angle of intersection of the circles x^(2)+y^(2)=4 and x^(2)+y^(2)+2x+2y , is

Find the radical center of the circles x^2+y^2+4x+6y=19 ,x^2+y^2=9,x^2+y^2-2x-4y=5,