JEE Main 2023 — 12 April, Morning Shift. Previous Year Question.
Problem
Two circles in the first quadrant of radii $r_1$ and $r_2$ touch the coordinate axes. Each of them cuts off an intercept of $2$ units with the line $x+y=2$. Then $r_1^2+r_2^2-r_1r_2$ is equal to ____.
Key insight. A circle in the first quadrant touching both axes must have its centre at $(r,r)$ with radius $r$ — a single unknown pins down the whole circle. Combining the perpendicular-distance-from-centre-to-line formula with the standard “chord length” formula for a circle gives one equation in $r$, which turns out to have exactly two valid solutions — the two radii $r_1$ and $r_2$.
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Approach
For a circle of radius $r$ touching both axes in the first quadrant, the centre is $(r,r)$. Find the perpendicular distance $d$ from this centre to the line $x+y=2$, then use the chord-length formula $\text{length}=2\sqrt{r^2-d^2}$ with the given intercept length of $2$. Solve the resulting equation for $r$ to find both radii, then compute $r_1^2+r_2^2-r_1r_2$.
Solution
Step 1 — Set up the circle’s centre
A circle of radius $r$ in the first quadrant touching both the x-axis and y-axis has centre $(r,r)$.
Step 2 — Find the perpendicular distance from the centre to the line
For $x+y-2=0$, the distance from $(r,r)$ is:
$$d = \frac{|r+r-2|}{\sqrt{1^2+1^2}} = \frac{|2r-2|}{\sqrt2}$$
Step 3 — Apply the chord-length formula
The intercept (chord) length on the line is given as $2$:
$$2\sqrt{r^2-d^2} = 2 \implies \sqrt{r^2-d^2}=1 \implies r^2-d^2=1$$
Step 4 — Substitute d²
$$d^2 = \frac{(2r-2)^2}{2} = 2(r-1)^2$$
$$r^2 – 2(r-1)^2 = 1$$
Step 5 — Expand and simplify
$$r^2 – 2(r^2-2r+1) = 1$$
$$r^2-2r^2+4r-2=1 \implies -r^2+4r-3=0 \implies r^2-4r+3=0$$
Step 6 — Solve for r
$$(r-1)(r-3)=0 \implies r=1 \text{ or } r=3$$
Both are valid positive radii, so $r_1=1$ and $r_2=3$ (in either order).
Step 7 — Compute r₁² + r₂² − r₁r₂
$$1^2+3^2-(1)(3) = 1+9-3 = 7$$
Answer
$$7$$
Common mistakes
- Assuming a circle touching the axes has centre $(r,0)$ or $(0,r)$ instead of $(r,r)$. Touching both the x-axis and y-axis simultaneously (not just one) is what forces both coordinates of the centre to equal the radius.
- Sign or expansion errors when simplifying $r^2-2(r-1)^2=1$. Expanding $(r-1)^2$ incorrectly, or mishandling the resulting minus sign in front of $2(r-1)^2$, easily flips the final quadratic’s coefficients and produces the wrong pair of radii.
Practise next
- Two circles in the first quadrant touch both coordinate axes, each cutting an intercept of $\sqrt2$ units on the line $x+y=3$. Find $r_1^2+r_2^2-r_1r_2$, using the same centre-distance-chord approach.
Show answer
$21$. A first-quadrant circle touching both axes has centre $(r,r)$ and radius $r$.
Its distance to $x+y=3$ is $\dfrac{|2r-3|}{\sqrt2}$, and a chord of length $\sqrt2$ has half-length $\tfrac{1}{\sqrt2}$, so
$$r^2-\frac{(2r-3)^2}{2}=\frac12\ \Longrightarrow\ r^2-6r+5=0,$$
giving $r_1=1$ and $r_2=5$. Then $r_1^2+r_2^2-r_1r_2=1+25-5=21$.
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