NIMCET 2018 Question Paper with solutions

To find the intersection point of circle, put x = 3 - 3y in equation of circle, we have
${\left(3-3\mathit{y}\right)}^2+{\mathit{y}}^2$ 10 (3 - 3y) - 12y +51=0
⇒ 9 + $9{\mathit{y}}^2$ - 18y + ${\mathit{y}}^2$ + 30 - 30y - 12y + 51 = 0
⇒ $10{\mathit{y}}^2$ - 60y + 90 = 0
⇒ ${\mathit{y}}^2$ - 6y + 9 = 0
⇒ ${\left(\mathit{y}-3\right)}^2$ = 0
⇒ y = 3
∴ x = 3 - 3y = 3 - 3 × 3 = - 6
∴ Intersection point is (-6, 3),