IPMAT Indore 2025 · Quantitative Ability · Question 26

Step 1 — what “touches the y-axis at \left( 0,4 \right)” tells us. At a point of tangency the radius is perpendicular to the line. The y-axis is vertical, so the radius to \left( 0,4 \right) is horizontal — meaning the centre sits at the same height:

\text{Centre}=\left( h,\,4 \right)

Also, the distance from the centre to the y-axis is the radius, so r=\left| h \right|.

Step 2 — use the second point. The circle passes through \left( -2,0 \right), so its distance from the centre is also r:

{\left( -2-h \right)}^{2}+{\left( 0-4 \right)}^{2}={r}^{2}={h}^{2}

Step 3 — expand and solve.

{h}^{2}+4h+4+16={h}^{2}\Rightarrow 4h+20=0 \Rightarrow h=-5

Step 4 — read off the radius.

r=\left| h \right|=5

(The centre is \left( -5,4 \right), to the left of the y-axis — which makes sense, since the circle must reach \left( -2,0 \right).)

Answer: Option 2 — 5