C2 2016 3 Edexcel

(a)

We have:

\begin{align*}|PQ| & = \sqrt {(13 - 8)^2 + (10 - 7)^2} \\ & = \sqrt {5^2 + 3^2} = \sqrt {25 + 9} \\ & = \sqrt {37}\end{align*}

(b)

The circle C then has equation:

(x - 7)^2 + (y - 8)^2 = (\sqrt {37})^2 = 37

The line l is perpendicular to PQ. PQ has gradient:

\displaystyle \frac {13 - 8} {10 - 7} = \frac 5 3

\displaystyle -\frac 3 5

Hence l has equation (since it has gradient -\dfrac 3 5 and passes through (10, 13)):

\displaystyle y - 13 = -\frac 3 5 (x - 10)

So:

\displaystyle 5y - 65 = -3x + 30

So:

\displaystyle 5y + 3x - 95 = 0

in particular a = 3, b = 5, c = -95.

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