Question

Let the circle $(x - 1)^2 + (y - 2)^2 = 25$ cuts rectangular hyperbola with transverse axis along y = x at 4 points A, B, C and D having coordinates $(x_i, y_i)$, $i = 1, 2, 3, 4$ such that origin is centre of hyperbola. If $\ell = x_1 + x_2 + x_3 + x_4$, $m = x_1^2 + x_2^2 + x_3^2 + x_4^2$, $n = y_1^2 + y_2^2 + y_3^2 + y_4^2$ then $\frac{m+n}{\ell} =$

• A. 100
• B. 10
• C. 50
• D. 20

Answer 1

Answer: (C)

50

Answer 2

The circle is

$$(x-1)^2+(y-2)^2=25,$$

and the rectangular hyperbola with center at the origin and transverse axis along $y=x$ can be written (after rotation) as

$$xy=d,$$

where $d$ is a constant.

For any intersection point $(x_i, y_i)$ we have

$$y_i=\frac{d}{x_i}.$$

Substitute $y=d/x$ into the circle:

$$(x-1)^2+\Big(\frac{d}{x}-2\Big)^2=25.$$

Multiplying by $x^2$ yields the quartic in $x$:

$$x^4-2x^3-20x^2-4dx+d^2=0.$$

Let the roots be $x_1,x_2,x_3,x_4$. By Vieta’s formulas:

• Sum of the roots: $$\ell=x_1+x_2+x_3+x_4=2.$$
• Sum of the product of the roots taken two at a time is $S_2=-20$.

Then the sum of squares of the $x$-coordinates is

$$m=\sum x_i^2=(\sum x_i)^2-2S_2=2^2-2(-20)=4+40=44.$$

For the corresponding $y$-coordinates, because $y_i=\frac{d}{x_i}$,

$$n=\sum y_i^2=d^2\sum \frac{1}{x_i^2}.$$

To compute $\sum 1/x_i^2$, first find $\sum 1/x_i$ from Vieta:

$$\sum \frac{1}{x_i}=\frac{x_2x_3x_4+\cdots}{x_1x_2x_3x_4}=\frac{S_3}{d^2}=\frac{4d}{d^2}=\frac{4}{d},$$

where $S_3=4d$ (using sign changes from the quartic).

Now, using the identity

$$\sum \frac{1}{x_i^2}=\left(\sum \frac{1}{x_i}\right)^2-2\sum_{i<j} \frac{1}{x_ix_j},$$

one can show (by writing the reciprocal polynomial and applying Vieta) that

$$\sum \frac{1}{x_i^2}=\frac{56}{d^2}.$$

Thus,

$$n=d^2\cdot\frac{56}{d^2}=56.$$

Therefore,

$$m+n=44+56=100,\quad\text{and}\quad\frac{m+n}{\ell}=\frac{100}{2}=50.$$