A line tangent to a circle

Question

Can anyone help me to solve this?

Determine the value or values of $k$ such that $x + y + k = 0$ is tangent to the circle $x^2+y^2+6x+2y+6=0$.

I don't know how to calculate the tangent.

Answer 1

Developed hint:

If the line is tangent to the circle then the line's distance to the circle's center equals the circle's radius. The circle's equation is

$$x^2+y^2+6x+2y+6=(x+3)^2-9+(y+1)^2-1+6\implies$$

$$(x+3)^2+(y+1)^2=4$$

Well, now use the formula for the distance of the point $\;(a,b)\;$ to the line $\;Ax+By+C=0\;$ , which is

$$\frac{|Aa+Bb+C|}{\sqrt{A^2+B^2}}$$

and this distance has to be equal to $\;2\;$ ...

Answer 2

Hints:

• A line is tangent to a circle if there exist precisely one point that is both on the straight line and on the circle
• A point $(x,y)$ is both on the line and the circle if it satisfies both equations.
• A quadratic equation has exactly one solution if its discriminant is $0$.

Answer 3

The tangent's gradient is $-1$. From $$(x+3)^2+(y+1)^2=4$$ we know that the normal's gradient is $(y+1)/(x+3)$, so we must have $(y+1)/(x+3)=1$, that is $y+1=x+3$, hence $2(x+3)^2=4$. We conclude $x=-3\pm\sqrt2$ and $y=x+2=-1\pm\sqrt2$.

So the points where the tangents touch the circle are $(-3\pm\sqrt2,-1\pm\sqrt2)$. From here is is well known (isn't it?) how to derive the tangent's equations: $$(x+3)(\pm\sqrt2)+(y+1)(\pm\sqrt2)=4.$$ Now expand and compare coefficients. No need for discriminants here.

Edit: In fact we do not have to know the coordinates of the two points, but solely $x+3$ and $ y+1$, which makes the calculation even simpler.