+6248 \(\text{l =AP will be perpendicular to OA as AP is tangent to the circle}\\ \text{the slope of OA is }m_{OA} = \dfrac{7-0}{1-0} = 7 \\ \text{The slope of AP is thus } m_{AP} = -\dfrac{1}{m_{OA}}=-\dfrac 1 7\)
\(\text{The line with slope }m_{AP} \text{ passing through point }A \text{ is given by}\\ (y-7) = -\dfrac 1 7(x-1) \\ y = \dfrac{-x + 50}{7} \\ \text{and this line intersects the x axis at} \\ 0=\dfrac{-x+50}{7} \Rightarrow x = 50\)
\(\text{so }P=(50,0) \\ |OA| = \sqrt{50} \text{ as this is the radius of the circle} \\ |AP| = \sqrt{(50-1)^2 + (0-7)^2} = \sqrt{2450} = 35\sqrt{2}\\ \text{As OAP is a right triangle, the area is given by}\\ area = \dfrac 1 2 b h = \dfrac 1 2|AP| |OA|= \dfrac 1 2 \sqrt{50}\cdot35\sqrt{2} = 175\)
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