MHTCET 18 Apr 2026 Shift 2 Paper

Concept:The centre of a circle that touches both given lines must lie on one of their angle bisectors.Explanation:Write the lines as $x+y-1=0$ and $x-y-1=0$.For a circle touching both lines, its centre is equidistant from them.So the centre lies on the angle bisectors:$$\frac{x+y-1}{\sqrt{2}} = \pm \frac{x-y-1}{\sqrt{2}}.$$This simplifies to the two loci: $y=0$ or $x=1$.Case 1: Centre on $y=0$, say $(h,0)$.Radius is the distance from $(h,0)$ to either line:$$r = \frac{|h-1|}{\sqrt{2}}.$$Since the circle passes through the origin, $r = |h|$.Therefore,$$|h| = \frac{|h-1|}{\sqrt{2}}.$$Squaring gives:$$2h^2 = (h-1)^2 \Rightarrow h^2 + 2h - 1 = 0.$$This quadratic has two distinct real roots:$$h = -1 \pm \sqrt{2}.$$Thus, two circles are possible from this case.Case 2: Centre on $x=1$, say $(1,k)$.Passing through the origin requires:$$\sqrt{1+k^2} = \frac{|k|}{\sqrt{2}}.$$Squaring gives $1+k^2 = \frac{k^2}{2}$, which has no real solution.So no circle is obtained from this case.Answer:The total number of circles is $2$.Hence, the correct option is B.