Find distance between intersections of circle and line?

[ijabbott]

(02-07-2019 03:48 PM)kevin3g Wrote:  The equations are y=2x+3 and (x+1)^2+y^2=55. I know you can put them in solve and find x and y twice, but is there a faster way?

I don't know if it is faster, but:

\[
y=2x+3, (x+1)^2+y^2=55 \\
(x+1)^2+(2x+3)^2=55 \\
x^2+2x+1+4x^2+12x+9=55 \\
5x^2+14x-45=0
\]

If the quadratic equation \(ax^2+bx+c=0\) has real roots, the roots are separated by \(\frac{\sqrt{b^2-4ac}}{a}\), since \(\frac{-b+\sqrt{b^2-4ac}}{2a}-\frac{-b-\sqrt{b^2-4ac}}{2a} = \frac{2\sqrt{b^2-4ac}}{2a} = \frac{\sqrt{b^2-4ac}}{a}\). This gives the \(x\) separation of the intersections, \(\Delta x = \frac{\sqrt{b^2-4ac}}{a}\).

The \(y\) separation, \(\Delta y\) will be twice that since \(y=2x+3\), so \(\Delta y = 2\Delta x\) (and the \(3\)s cancel). The distance between the intersection points \(D = \sqrt{(\Delta x)^2 + (\Delta y)^2} = \sqrt{(\Delta x)^2 + (2\Delta x)^2} = \sqrt{5(\Delta x)^2} = \sqrt{5\left(\frac{\sqrt{b^2-4ac}}{a}\right)^2} =\sqrt{5\left(\frac{b^2-4ac}{a^2}\right)} = \frac{\sqrt{5(b^2-4ac)}}{a}\).

Plugging in the coefficients of the quadratic equation, \(D = \frac{\sqrt{5(14^2- (4 \cdot 5 \cdot (-45)))}}{5} = \frac{\sqrt{5(196+900)}}{5} = \frac{\sqrt{5480}}{5} \approx 14.81\).

EDIT: Corrected mistake in \(c\) spotted by Albert Chan. \(c\) should be \(-45\), not \(-41\).