JEE Mathematics 2023 Question with Solution

Given: Two circles in the first quadrant touch both coordinate axes, so their centres are of the form $(a,a)$ and radius is $a$.

Find: The value of $r_1^2 + r_2^2 - r_1 r_2$ when each circle cuts off an intercept of $2$ units on the line $$ x+y=2

$$$$

The equation of such a circle is

$$(x-a)^2+(y-a)^2=a^2$$

Expanding,

$$x^2+y^2-2ax-2ay+a^2=0$$

The chord cut by the line $x+y=2$ has length $2$. For a circle of radius $a$, chord length is

$$2\sqrt{a^2-d^2}=2$$

where $d$ is the perpendicular distance from the centre $$ (a,a)

$$$$

Now,

$$d=\frac{|a+a-2|}{\sqrt{2}}=\frac{|2a-2|}{\sqrt{2}}$$

Substituting into the chord formula,

$$2\sqrt{a^2-\left(\frac{2a-2}{\sqrt{2}}\right)^2}=2$$

Squaring both sides,

$$a^2-\frac{(2a-2)^2}{2}=1$$

Simplifying,

$$2a^2-(2a-2)^2=2$$ $$2a^2-(4a^2-8a+4)=2$$ $$-2a^2+8a-4=2$$ $$-2a^2+8a-6=0$$

Dividing by $-2$,

$$a^2-4a+3=0$$

Thus,

$$(a-1)(a-3)=0$$

So the two possible radii are

$$r_1=1,\quad r_2=3$$

Therefore,

$$r_1+r_2=4,\quad r_1r_2=3$$

Hence,

$$r_1^2+r_2^2-r_1r_2=(r_1+r_2)^2-3r_1r_2=4^2-3\cdot 3=16-9=7$$

Therefore, the required value is $7$.

Given: The circle touches both axes, so its centre is $(r,r)$ and radius is $r$.

Find: The possible radii and then compute $$ r_1^2+r_2^2-r_1r_2

$$$$

From the geometry shown, half of the intercepted chord is taken as $AB=1$, so in the right triangle,

$$OA=\sqrt{r^2-1}$$

The perpendicular distance from $(r,r)$ to the line $$ x+y=2

$$$$

\frac{|2r-2|}{\sqrt{2}}

$$Hence,$$

\frac{|2r-2|}{\sqrt{2}}=\sqrt{r^2-1}

$$Squaring,$$

2(r-1)^2=r^2-1

$$$$

2r^2-4r+2=r^2-1

$$$$

r^2-4r+3=0

$$So,$$

(r-1)(r-3)=0

Thus the radii are $$ 1 $$ and $$ 3 $$. Therefore,

r_1^2+r_2^2-r_1r_2=1+9-3=7

So the required value is **$$7$$**.