The equation of circle passing through $(0,0)$ and points of intersection of the circle $x^2 + y^2 + 2x + 4y -4 = 0$ and $y - 7x = 2$

Question

I have recently been given this question, and have been puzzling over it for days.

Find the equation of the circle passing through the origin and the points of intersection of the circle $x^2 + y^2 + 2x + 4y -4 = 0$ and the line $y - 7x = 2$.

Even stranger, the solution goes like so:
Let $(x^2 + y^2 + 2x + 4y - 4) + k(y - 7x - 2) = 0 \;\;\;\;\; ...\;(1)$
Since (1) passes through the origin,
$k = -2$
Substituting into (1) and simplifying,
$x^2 + y^2 + 16x +2y = 0$

Yet I do not understand why $k=-2$ because it passes through the origin.

Could someone provide an alternative solution or explain the one provided please?

Answer 1

The equation $$(x^2 + y^2 + 2x + 4y - 4) + k(y - 7x - 2) = 0 \tag1$$ is equivalent to $$ x^2 + y^2 + (2 - 7k)x + (4 + k)y - (4 + 2k) = 0,$$ which is clearly the equation of a circle. Moreover, if a point $P = (x_P, y_P)$ is both on the circle $x^2 + y^2 + 2x + 4y - 4 = 0$ and on the line $y - 7x - 2 = 0,$ then $$(x_P^2 + y_P^2 + 2x_P + 4y_P - 4) + k(y_P - 7x_P - 2) = 0 + k(0) = 0, $$ so $P$ is on the circle described by Equation $(1)$. The same is true for the other point of intersection of the given circle line.

So now you have an equation of a circle that goes through the two intersection points. If it also goes through the origin, then it is the circle you want. At the origin, $x = y = 0,$ so Equation $(1)$ becomes $$(0^2 + 0^2 + 2(0) + 4(0) - 4) + k(0 - 7(0) - 2) = -4 - 2k = 0,$$ which is possible only if $k = -2.$

Answer 2

"passes through the origin" means that $(x,y)=(0,0)$ satisfies the equation (1). Substitute those values in (1), and you get $$(-4)+k(-2)=0$$ $$k=-2$$

Answer 3

The conditions are

$$ x^2+y^2+2x+4y-4=0\\ x^2+y^2-2x_0 x-2y_0y = 0\\ y-7x-2=0 $$

subtracting the two circle equations

$$ 2(x_0+1)x+2(y_0+2)y = 4 $$

comparing with the last

$$ (x_0+1)x+(y_0+2)y = 2\\ y-7x = 2 $$

we obtain

$$ x_0 = -8, y_0 = -1 $$