Question

Prove that the straight line is tangent to the circle?

Answer 1

Determine the derivative:

`2x + 2y(dy/dx) = 0`

`2y(dy/dx) = -2x`

`dy/dx = -x/y`

The straight line with equation `x = 2y + 5` has slope of `1/2`.

`1/2 = -x/y`

`-1/2 = x/y`

`y = -2x`

Our second equation is `x^2 + y^2 = 5`. Therefore:

`(-2x)^2 + x^2 = 5`

`5x^2 = 5`

`x = +- 1`

But the tangent line at `x= -1` doesn't have a y-interept of `-5/2`, therefore, `x = 1`.

Thus, there is indeed a point on the circle where `x =2y + 5` is tangent to `x^2 + y^2 = 5`.

Hopefully this helps!

Answer 2

If you need to prove it algebraically, I understand this means you should not use derivatives.

Note then that if and only if the straight line:

`x =2y+5`

is tangent to the circle

`x^2+y^2 =5`

then they must have a single point in common, whose coordinates solve the system:

`{( x -2y -5 = 0),(x^2+y^2-5 =0):}`

Substitute `x` from the first equation into the second:

`(2y+5)^2+y^2 - 5 = 0`

`4y^2+20y+25+y^2 -5 =0`

`5y^2+20y+20`

`y^2 +4y+4 =0`

For which in facct the determinant is null, because:

`(y+2)^2 =0`

and the only solution is `y=-2` and consequently `x = 1`

The point `P(1,-2)` is then the only point in common between the line and the circle and this means that the line is tangent to the circle.