Find the points of intersection of the line $x + 16 = 7y$ and the circle $x^2 + y^2 - 4x + 2y = 20$

Question

I have the following problem:

On a single set of coordinate axes, sketch the line $x + 16 = 7y$ and the circle $x^2 + y^2 - 4x + 2y = 20$ and find their points of intersection.

So far I have only the center of the circle and its radius, I already tried to eliminate $x$ to solve with $y$ without success, any clues?

Edit:

I already tried with $(7y - 16)^2 + y^2 - 4(7y - 16) + 2y - 20 = 0$ but the result is wrong, I checked what could be the mistake but I don't see it, maybe I skipped something with the algebra process.

Answer 1

Solve for $x$ in the equation of the line: $$x=7y-16$$ Then plug your new equation for $x$ into the equation of the circle: $$\left(7y-16\right)^2+y^2-4\left(7y-16\right)+2y-20=y^2-5y+6=0$$ This is a quadratic in terms of $y$; the solutions to the equation are $y=2$ and $y=3$. Since you said that you couldn't find the correct answer using this formula, here is the process to solve this using the quadratic formula: $$y=\frac{5\pm\sqrt{25-4\left(1\right)\left(6\right)}}{2}$$ $$y=\frac{5\pm1}{2}$$ $$y=2,y=3$$ You can plug these values back into the equation of the line to find the $x$-values at which the circle and line intersect; these will be $x=-2$ and $x=5$. Therefore, the points of intersection are $\left(-2,2\right)$ and $\left(5,3\right)$. We can check this graphically as well:

Answer 2

HINT

Yes we need to substitute $x$ (or $y$) from the line equation into the circle equation and solve the quadratic equation obtained in $y$ (or $x$), that is

$$(7y-16)^2+y^2-4(7y-16)+2y-20=0\implies y^2-5y+6=0$$

Answer 3

Since you asked for another way to solve it, here’s a method for computing the intersection of a line and any conic. It’s overkill for a circle-line intersection, though.

The general equation $ax+by+c=0$ of a line can be written as $(a,b,c)^T\cdot(x,y,1)^T=0$, so a line can be represented by the homogeneous vector $\mathbf l = [a,b,c]^T$, with every point $\mathbf x$ on the line satisfying $\mathbf l^T\mathbf x = 0$. Similarly, the equation of any conic can be written as $\mathbf x^TC\mathbf x=0$, where $C$ is a symmetric matrix. The matrix $[\mathbf l]_\times^TC[\mathbf l]_\times$, where $[\mathbf l]_\times$ is the “cross-product matrix” of $\mathbf l$—the skew-symmetric matrix such that $[\mathbf l]_\times\mathbf x = \mathbf l\times\mathbf x$—turns out to be a degenerate dual conic matrix that represents that point of intersection of the line and conic. That is, it is some scalar multiple of $\mathbf p\mathbf q^T+\mathbf q\mathbf p^T$, where $\mathbf p$ and $\mathbf q$ are the points of intersection. We “split” this dual conic by finding a value of $\alpha$ such that $[\mathbf l]_\times^TC[\mathbf l]_\times+\alpha[\mathbf l]_\times$ is a rank-one matrix, from which we will be able to read $\mathbf p$ and $\mathbf q$ directly because it is a multiple of $\mathbf p\mathbf q^T$.

For your problem, $$\mathbf l = \begin{bmatrix}1\\-7\\16\end{bmatrix}, [\mathbf l]_\times = \begin{bmatrix}0&-16&-7\\16&0&-1\\7&1&0\end{bmatrix} \\ C = \begin{bmatrix}1&0&-2\\0&1&1\\-2&1&-20\end{bmatrix} \\ [\mathbf l]_\times^TC[\mathbf l]_\times+\alpha[\mathbf l]_\times = \begin{bmatrix}-500 & 100-16\alpha & 75-7\alpha \\ 100+16\alpha & 300 & 125-\alpha \\ 75+7\alpha & 125+\alpha & 50 \end{bmatrix}.$$ This last matrix will have rank one if all of the $2\times2$ minors vanish. Choosing the lower-right produces the quadratic equation $\alpha^2-625=0$, so $\alpha=\pm25$. Choosing the negative root yields $$\begin{bmatrix}-500&500&250\\-300&300&150\\-100&100&50\end{bmatrix}.$$ We can take any row/column pair with a nonzero diagonal entry. Taking the last, we have $\mathbf p = [250,150,50]^T$ and $\mathbf q = [-100,100,50]^T$, which dehomogenize to $(5,3)$ and $(-2,2)$.