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Here is a circle, centre O, and the tangent to the circle at the
point (8, 15) on the circle.
x
P (8, 15)
-17
O
17
Find an equation of the tangent at the point P.
Give your answer in the form y = ax + b where a and b are both
fractions with denominator 15.​

Here is a circle centre O and the tangent to the circle at thepoint 8 15 on the circlexP 8 1517O17Find an equation of the tangent at the point PGive your answer class=

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Answer:

The equation of the tangent is [tex]y=-\frac{8}{15}x+\frac{289}{15}[/tex].

Step-by-step explanation:

Center = (0, 0). Point on the circle = (8, 15).

OP is a radius. The slope of the radius = [tex]\frac{15-0}{8-0}=\frac{15}{8}[/tex].

The tangent is perpendicular to the radius.

So, its slope is = [tex]-\frac{8}{15}[/tex].

The tangent passes through (8, 15).

So, the equation is:

[tex]y-15=-\frac{8}{15}\left(x-8\right)\\y-15=-\frac{8}{15}x+\frac{64}{15}\\y=-\frac{8}{15}x+\frac{289}{15}[/tex]

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