Let us rewrite the equation of circle in form of
x2−8x+y2+2y+13=0,(x−4)2−16+(y+1)2−1+13=0,(x−4)2+(y+1)2=22.
Therefore, the center is located at (4,-1) and the radius is 2.
Let us substitute the coordinates of D in the equation obtained above.
(6−4)2+(−1+1)2=4,22+02=4.
The equality is true, so D lies on the circle.
The equation of the tangent line to the circumference at point (x1,y1) with center at (a,b) and radius r
is (x1−a)(x−a)+(y1−b)(y−b)=r2.
In our case
(6−4)(x−4)+(−1−(−1))(y−(−1))=22,2(x−4)=4,x−4=2,x=6.
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