Show that the lines joining the origin to the point of intersection of the line Show that the lines joining the origin to the point of intersection of the line $x+y=1$ with the curve $4x^2+4y^2+4x-2y-5=0$ are at right angles to each other. 
How do I approach this? Please help.
 A: By the process of homogenisation the pair of lines have the equation
$4(x^2+y^2)+(4x-2y)(x+y)-5(x+y)^2=0$ which simplifies to $3x^2-8xy-3y^2=0$
We know that for the pair of lines $ax^2+2hxy+by^2=0, a+b = 0 \Rightarrow$ they are orthogonal.
Here the same holds, and hence the pair of lines are perpendicular.
A: Hint:
The equation of any quadratic curve passing through the intersection will be $$4x^2+4y^2-2y+4x+5+k(x+y-1)^2=0$$
Now this has to pass through the origin $\implies5+k=0$
Now rearrange to form a quadratic equation in $x$ or in $y$
and solve.
A: Solving of the system gives 
$$4x^2+4(1-x)^2+4x-2(1-x)-5=0$$ or
$$8x^2-2x-3=0,$$ which gives the following points: $\left(\frac{3}{4},\frac{1}{4}\right)$ and $\left(-\frac{1}{2},\frac{3}{2}\right)$.
Now, $$m_1=\frac{\frac{1}{4}}{\frac{3}{4}}=\frac{1}{3}$$ and
$$m_2=\frac{\frac{3}{2}}{-\frac{1}{2}}=-3$$ and since $m_1m_2=-1$, we are done!
A: Hint:
Replace $x$ with $1-y$ in the equation of the circle to find the two abcissa of the intersection.
Now we know how to find the gradient of a straight line joining two points
