l =AP will be perpendicular to OA as AP is tangent to the circlethe slope of OA is mOA=7−01−0=7The slope of AP is thus mAP=−1mOA=−17
The line with slope mAP passing through point A is given by(y−7)=−17(x−1)y=−x+507and this line intersects the x axis at0=−x+507⇒x=50
so P=(50,0)|OA|=√50 as this is the radius of the circle|AP|=√(50−1)2+(0−7)2=√2450=35√2As OAP is a right triangle, the area is given byarea=12bh=12|AP||OA|=12√50⋅35√2=175
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