Question

I would like to know what formula I would use to solve this questionCircle P has a radius of 5cm. Point Q lies on circle P, and the equation of PQ is y=2/5x+3. Which equation could be the equation of a line tangent to circle P at point Q?

Answer 1

The line PQ and the line tangent to PQ at point Q are perpendicular lines, since P is the circle center and Q is on the circle.

Perpendicular lines have the following relation about their slopes:

`m_2=-(1)/(m_1)`

Comparing the equation of line PQ with the slope-intercept form of the linear equation (y = mx + b), we have a slope m = 2/5.

Therefore the slope of the tangent line is:

`m_2=-(1)/((2)/(5))=-(5)/(2)`

Now, let's put the equation in the standard form:

`\begin{gathered} y=-(5)/(2)x+b \\ y+(5)/(2)x=b \\ 2y+5x=2b \\ 5x+2y=2b \end{gathered}`

The only option with "5x + 2y" on the left side is the third option.