The line L is a tangent to the circle x2 + y2 = 45 at the point (-3, 6).
The line L crosses the x-axis at the point P.
Work out the coordinates of P.

Answer:

The point P is (-15,0).

Step-by-step explanation:

We need to find the slope of the line L to get the coordinates of P.

We know that L is tangent to the circle at point (-3, 6), which means L is perpendicular to the radius of the circle (a line between the origin (0,0) and the tangent point (-3,6))

So let's find the slope or the radius.

[tex]m_{R}=\frac{6-0}{-3-0}[/tex]    

[tex]m_{R}=-2[/tex]

Now, we know that if two lines are perpendiculars their slopes follow the rule:

[tex]m_{L}=-\frac{1}{m_{R}}[/tex]  

[tex]m_{L}=\frac{1}{2}[/tex]  

Knowing the slope of L we can find the equation of this line. The slope-intercept equation of a line is:

[tex]y=mx+b[/tex]

[tex]y=\frac{1}{2}x+b[/tex]

Using the point (-3,6) we can find b.

[tex]6=\frac{1}{2}(-3)+b[/tex]

[tex]b=\frac{15}{2}[/tex]

The line equation will be:

[tex]y=\frac{1}{2}x+\frac{15}{2}[/tex]

The point P has only an x-coordinate value (x,0)

So we have:

[tex]0=\frac{1}{2}x+\frac{15}{2}[/tex]

[tex]x=-15[/tex]

Therefore, the point P is (-15,0).

I hope it helps you!

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