If the lines `2x+3y+1=0` and `3x-y-4=0` lie along diameter of a circle of circumference `10pi`, then show that the equation of the circle is `x^(2)-y^(2)-2x+2y-23=0`

If the lines 2x+3y+1=0 and 3x-y-4=0 lie along diameters of a circle of circumference 10 pi , then the equation of the circle is

If the lines x+y+3=0 and 2x-y-9=0 lie along diameters of a circle of circumference 8pi then the equation of the circle is :

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 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 3x-4y-7 = 0 and 2x-3y-5=0 are two diameters of a circle of area 49pi square units, the equation of the circle is:

Find the parametric equation of the circles : 3x^2 + 3y^2 + 4x-6y - 4 = 0

The equation of the normal to the circle x^(2)+y^(2)+6x+4y-3=0 at (1,-2) to is

Find the equation of image circle of the circle x^(2)+y^(2)-2x=0 in the line x+y-2=0

The circumference of circle x^(2)+y^(2)-10y-36=0 is

The line 4x+3y-4=0 divides the circumference of the circle centred at (5,3) in the ratio 1:2. Then the equation of the circle is