Dear Uncle Colin,

How would I show that $4y - 3x + 26 = 0$ is a tangent to $(x+4)^2 +(y-3)^2 = 100$?

Can I Reconcile Circle/Line Equations?

Hi, CIRCLE, and thanks for your message!

The simplest way (I think) is to note that if the line is tangent to the circle, solving the equations simultaneously will give a single, repeated root.

Let’s solve that. I’ll start by replacing the $x+4$ in the second equation.

• $x = \frac{4y + 26}{3}$
• $x+4 = \frac{4y + 38}{3}$
• $(x+4)^2 = \frac{16y^2 + 304y + 1444}{9}$

So $\frac{16y^2 + 304y + 1444}{9} + (y^2 - 6y + 9) = 100$

I’m not a fan of the fractions, so let’s multiply by 9:

• $16y^2 + 304y + 1444 + 9y^2 - 54y + 81 = 900$

… and rearrange:

• $25y^2 + 250y -625 = 0$

Hey lookit! A nice common factor!

• $y^2 + 10y - 25 = 0$

And that factorises nicely as $(y - 5)^2 = 0$, giving a double root. (Of course, we could have used the discriminant a little sooner.)

An alternative method is to find the radius of the circle perpendicular to the line. The original line has a gradient of $\frac{3}{4}$, so the radius must have a gradient of $-\frac{4}{3}$, and pass through $(-4,3)$.

The equation of that line is $y-3 = -\frac{4}{3}(x+4)$, or $4x + 3y + 7 = 0$.

Where does that meet the original line?

• Radius: $4x + 3y + 7 = 0$ so $-3y = 4x + 7$
• Original: $-3x + 4y + 26 =0$, so $4y = 3x - 26$

• Multiply the radius by -4: $12y = -16x - 28$
• Multiply the original by 3: $12y = 9x - 78$

• So $-16x - 28 = 9x - 78$, so $50 = 25x$ and $x=2$.

That gives $y=-5$, so thee radius crosses the line at $(2,-5)$.

That is a distance of $\sqrt{6^2 + 8^2} = 10$ from the centre, so the original line is perpendicular to the radius and meets it on the circumference – and must be a tangent!

Hope that helps!

- Uncle Colin