MHTCET 11 Apr 2026 Shift 1 Paper

Concept:The centre of a circle always lies on the perpendicular bisector of any chord of that circle. The two given points form a chord, so the centre is the intersection of this perpendicular bisector and the given line.Explanation:Let the given points be $A(2,3)$ and $B(4,5)$.The midpoint of $AB$ is $\left(\frac{2+4}{2}, \frac{3+5}{2}\right) = (3,4)$.The slope of $AB$ is $\frac{5-3}{4-2} = 1$.Therefore, the perpendicular bisector has slope $-1$ and passes through $(3,4)$:$y - 4 = -1(x - 3)$$\Rightarrow y = -x + 7$.The centre also satisfies the given line $y - 4x + 3 = 0$, which can be written as $y = 4x - 3$.Equating the two expressions for $y$:$-x + 7 = 4x - 3$$\Rightarrow 5x = 10$$\Rightarrow x = 2$.Substituting $x = 2$ into $y = 4x - 3$ gives $y = 4(2) - 3 = 5$.Thus, the centre is $(2,5)$.The radius is the distance from the centre $(2,5)$ to the point $(2,3)$:$r = |5 - 3| = 2$.So the equation of the circle is:$(x - 2)^2 + (y - 5)^2 = 2^2$$\Rightarrow x^2 - 4x + 4 + y^2 - 10y + 25 = 4$$\Rightarrow x^2 + y^2 - 4x - 10y + 25 = 0$.Answer:The required equation is $x^{2} + y^{2} - 4x - 10y + 25 = 0$, which is Option A.