Find an unknown coefficient in a line equation...

Question

So, I have to find the unknown coefficient in this line: $$x+y+C=0$$ so that it is a tangent to this circle: $$x^2+y^2-5x-7y+6=0$$ I've transformed the circle equation to this form: $$(x-\frac{5}{2})^2+(y-\frac{7}{2})^2=\frac{25}{2}$$ But I can't see how it can help me lol :D

Answer 1

Substitute $y = C-x$ into the equation of the circle to get:

$x^2+(C-x)^2-5x-7(C-x)+6 = 0$

$2x^2+(2-2C)x+(C^2-7C+6) = 0$

If the line is tangent to the circle, this equation has only one solution for $x$. So the discriminant of the quadratic should be $0$, i.e.

$(2-2C)^2 - 4 \cdot 2 \cdot (C^2-7C+6) = 0$

This will give you two solutions for $C$. ($C=1$ and $C=11$)

Answer 2

Here is another way to view the problem. The by a change of coordinates we convert the problem to $$x^2+y^2=\frac{25}{2}$$ with the line $x+y+D=0$. You can easily find the relation between $D$ and $C$. Anyway the line $x+y+D$ has slope $-1$, so at the point of tangency the slope of the radius is $1$. That means this point occurs when $x=y$ as solving we have $x=\pm \frac{5}{2}$ which gives $D=\pm 5$, then $C$ can be found.