Question -
Answer -
It is given that thecircle has the radius 10 and has diameters 2x + y = 6 and 3x + 2y = 4.
We know that thecentre is the intersection point of the diameters.
On solving thediameters, we get the centre to be (8, -10).
We have a circle withcentre (8, -10) and having radius 10.
By using the formula,
We know that theequation of the circle with centre (p, q) and having radius ‘r’ is given by: (x– p)2 + (y – q)2 = r2
Where, p = 8, q = -10,r = 10
Now by substitutingthe values in the equation, we get
(x – 8)2 +(y – (-10))2 = 102
(x – 8)2 +(y + 10)2 = 100
x2 –16x + 64 + y2 + 20y + 100 = 100
x2 + y2 –16x + 20y + 64 = 0.
∴ The equation of thecircle is x2 + y2 – 16x + 20y + 64 = 0.