Show that the straight line with equation $4x+3y=25$ is a tangent to the circle with equation ${x}^{2}+{y}^{2}=25$.

-oOo-

### Solution.

There are two ways in which a circle and a line might intersect.

They might also not intersect at all. A circle and a line might intersect once or twice.

A singular intersection can only occur if the line is a tangent, so for this problem we need only show that there is one intersection.

The equation for the line can be re-written as

$y=\frac{25-4x}{3}$

Substituting this for y in the equation of the circle,

${x}^{2}+{\left(\frac{25-4x}{3}\right)}^{2}=25$

Squaring out the bracket and multiplying by nine,

$9{x}^{2}+{(25-4x)}^{2}=225$

$9{x}^{2}+625-200x+16{x}^{2}=225$

$25{x}^{2}-200x+400=0$

${x}^{2}-8x+16=0$

Using the determinant on this quadratic,

${b}^{2}-4ac={8}^{2}-4\times 1\times \mathrm{16}=0$

This shows there is only one intersection, so the line is a tangent to the circle.

-oOo-