Question

Two curves $x^2 + 2y^2 + 2xy - 2x - 2 = 0$ & $x^2 - y^2 + 2xy - 2y - 1 = 0$ intersects at $A_i$ for i = 1, 2, 3, 4.

If $L_i: x = m_i y$ denotes line $OA_i$ where O is origin and i = 1, 2, 3 & 4. Find the equation of circle with centre $(3\Sigma m_1m_2, 3\Sigma m_1m_2m_3)$ & radius $3\Sigma m_1$.

• A. $x^2 + y^2 + 32x - 64y + 1216 = 0$
• B. $x^2 + y^2 + 32x + 64y - 1216 = 0$
• C. $x^2 + y^2 + 32x + 64y + 1216 = 0$
• D. $x^2 + y^2 + 32x - 64y - 1216 = 0$

Answer 1

Answer: (C)

C

Answer 2

The given curves are: $C_1: x^2 + 2y^2 + 2xy - 2x - 2 = 0$ $C_2: x^2 - y^2 + 2xy - 2y - 1 = 0$

The lines $OA_i$ pass through the origin. Let $x = my$ be a line from the origin. Substitute $x=my$ into $C_1$ and $C_2$: For $C_1$: $(my)^2 + 2y^2 + 2(my)y - 2(my) - 2 = 0$ $m^2y^2 + 2y^2 + 2my^2 - 2my - 2 = 0$ $(m^2+2m+2)y^2 - 2my - 2 = 0$ (1)

For $C_2$: $(my)^2 - y^2 + 2(my)y - 2y - 1 = 0$ $m^2y^2 - y^2 + 2my^2 - 2y - 1 = 0$ $(m^2+2m-1)y^2 - 2y - 1 = 0$ (2)

Let $A = m^2+2m+2$ and $B = m^2+2m-1$. The system of equations is: $Ay^2 - 2my - 2 = 0$ $By^2 - 2y - 1 = 0$

To eliminate $y$, we can use the method of resultants or solve for $y$ from one equation and substitute into the other. From (2), $1 = By^2 - 2y$. Substitute into (1): $Ay^2 - 2my - 2(By^2 - 2y) = 0$ $Ay^2 - 2my - 2By^2 + 4y = 0$ $(A-2B)y^2 + (4-2m)y = 0$ Since $y \neq 0$ for the intersection points $A_i$ (as $y=0$ implies $x=0$, which is the origin, not an intersection point $A_i$), we can divide by $y$: $(A-2B)y + (4-2m) = 0$ $y = \frac{2m-4}{A-2B}$

Now substitute $A$ and $B$ back: $A-2B = (m^2+2m+2) - 2(m^2+2m-1) = m^2+2m+2 - 2m^2-4m+2 = -m^2-2m+4$. So, $y = \frac{2m-4}{-m^2-2m+4}$.

Substitute this expression for $y$ back into equation (2): $B \left(\frac{2m-4}{-m^2-2m+4}\right)^2 - 2 \left(\frac{2m-4}{-m^2-2m+4}\right) - 1 = 0$ $B \frac{4(m-2)^2}{(m^2+2m-4)^2} + \frac{4(m-2)}{m^2+2m-4} - 1 = 0$ Multiply the entire equation by $(m^2+2m-4)^2$: $4B(m-2)^2 + 4(m-2)(m^2+2m-4) - (m^2+2m-4)^2 = 0$ Let $P(m) = m^2+2m-4$. Then $B = m^2+2m-1 = P(m)+3$. $4(P(m)+3)(m-2)^2 + 4(m-2)P(m) - P(m)^2 = 0$ $4P(m)(m-2)^2 + 12(m-2)^2 + 4(m-2)P(m) - P(m)^2 = 0$ $P(m)[4(m-2)^2 + 4(m-2) - P(m)] + 12(m-2)^2 = 0$ $P(m)[4(m^2-4m+4) + 4m-8 - (m^2+2m-4)] + 12(m^2-4m+4) = 0$ $P(m)[4m^2-16m+16 + 4m-8 - m^2-2m+4] + 12m^2-48m+48 = 0$ $P(m)[3m^2-14m+12] + 12m^2-48m+48 = 0$ Substitute $P(m) = m^2+2m-4$: $(m^2+2m-4)(3m^2-14m+12) + 12m^2-48m+48 = 0$ Expand the product: $3m^4 - 14m^3 + 12m^2 + 6m^3 - 28m^2 + 24m - 12m^2 + 56m - 48 + 12m^2 - 48m + 48 = 0$ Combine like terms: $3m^4 + (-14+6)m^3 + (12-28-12+12)m^2 + (24+56-48)m + (-48+48) = 0$ $3m^4 - 8m^3 - 16m^2 + 32m = 0$ Factor out $m$: $m(3m^3 - 8m^2 - 16m + 32) = 0$ The four roots of this equation are $m_1, m_2, m_3, m_4$. One root is $m_1 = 0$. The other three roots are from the cubic equation $3m^3 - 8m^2 - 16m + 32 = 0$. Let these be $m_2, m_3, m_4$.

Using Vieta's formulas for $3m^3 - 8m^2 - 16m + 32 = 0$: Sum of roots: $m_2+m_3+m_4 = -(-8)/3 = 8/3$. Sum of products of roots taken two at a time: $m_2m_3+m_2m_4+m_3m_4 = -16/3$. Product of roots: $m_2m_3m_4 = -(32)/3 = -32/3$.

Now calculate the required sums involving all four $m_i$: $\Sigma m_1 = m_1 + (m_2+m_3+m_4) = 0 + 8/3 = 8/3$. $\Sigma m_1m_2 = m_1(m_2+m_3+m_4) + (m_2m_3+m_2m_4+m_3m_4) = 0 + (-16/3) = -16/3$. $\Sigma m_1m_2m_3 = m_1(m_2m_3+m_2m_4+m_3m_4) + m_2m_3m_4 = 0 + (-32/3) = -32/3$.

The center of the circle is $(h, k) = (3\Sigma m_1m_2, 3\Sigma m_1m_2m_3)$. $h = 3 \times (-16/3) = -16$. $k = 3 \times (-32/3) = -32$. So, the center is $(-16, -32)$.

The radius of the circle is $R = 3\Sigma m_1$. $R = 3 \times (8/3) = 8$.

The equation of the circle is $(x-h)^2 + (y-k)^2 = R^2$: $(x - (-16))^2 + (y - (-32))^2 = 8^2$ $(x+16)^2 + (y+32)^2 = 64$ $x^2 + 32x + 256 + y^2 + 64y + 1024 = 64$ $x^2 + y^2 + 32x + 64y + 256 + 1024 - 64 = 0$ $x^2 + y^2 + 32x + 64y + 1216 = 0$

This matches option (C).