UPSC NDA Math Model Paper 9

Given diameters' equation:
y + 2x - 3 = 0 ...(i)
2y + 3x - 2 = 0 ...(ii)
Substracting (ii) from 2 × (i)
x - 4 = 0
⇒ x = 4
From equation (i)
y + 2(4) - 3 = 0
⇒ y + 5 = 0
⇒ y = -5
∴ Coordinates of the centre are (4, -5)
Given the circle passes through (2, 2)
Now the radius r=√(y2−y1)2+(x2−x1)2
⇒r=√(−5−2)2+(4−2)2
⇒r=√49+4
⇒r2=53
∴ The equation of circle is:
(x−h)2+(y−k)2=R2
⇒(x−4)2+(y+5)2=53
⇒x2+y2−8x+10y+16 +25=53
⇒x2+y2−8x+10y−12=0