Question

I need help with this

Answer 1

A tangent to a circle is a straight line which touches the circle at only one point. Therefore, if
`x=2y+5` is a tangent of
`x^2+y^2=5` then there will be one point of intersection.

To find the point of intersection, substitute
`x=2y+5` into
`x^2+y^2=5` and solve for y:

`\implies (2y+5)^2+y^2=5`

`\implies 4y^2+20y+25+y^2=5`

`\implies 5y^2+20y+20=0`

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

`\implies (y+2)^2=0`

`\implies y+2=0`

`\implies y=-2`

Substitute found value of y into
`x=2y+5` and solve for x:

`\implies x=2(-2)+5=1`

Therefore, there is one point of intersection at (1, -2), thus proving that the straight line equation is a tangent to the circle.