Given that we need to find the equation of the circle which has diameters 2x - 3y = 5 and 3x - 4y = 7 and having area 154 square units.
We know that the centre of the circle is the point of intersection of the diameters.
On solving the diameters, we get the centre to be (1, - 1).
We know that the area of the circle is given by r2, where r is the radius of the circle.
⇒ r = 7
We have a circle with centre (1, - 1) and having radius 7 units.
We know that the equation of the circle with centre (p, q) and having radius ‘r’ is given by:
⇒ (x - p)2 + (y - q)2 = r2
Now we substitute the corresponding values in the equation:
⇒ (x - 1)2 + (y - ( - 1))2 = 72
⇒ (x - 1)2 + (y + 1)2 = 49
⇒ x2 - 2x + 1 + y2 + 2y + 1 = 49
⇒ x2 + y2 - 2x + 2y - 47 = 0
∴The equations of the circles is x2 + y2 - 2x + 2y - 47 = 0.
Rate this question :
If the lineRD Sharma - Mathematics
The sides of a sqRD Sharma - Mathematics
The line 2xRD Sharma - Mathematics
ABCD is a sRD Sharma - Mathematics
The circleRD Sharma - Mathematics