𝐶 is a circle represented by 𝑥² + 𝑦²﹣8𝑥﹣8𝑦 + 16 = 0. The straight line joining the origin to the center of the circle is extended such that it intersects the circle at a point (𝑝,𝑞). If a tangent to the circle is drawn at (𝑝,𝑞), then what is the area, in sq. units, of the triangle formed by the tangent with the coordinate axes?

The circle is (x-4)^2 + (y-4)^2 = 16. Center is (4,4), radius is 4. The line from origin to center is y=x. Intersection points (p,q) are found by substituting y=x into the circle equation: 2x^2 - 16x + 16 = 0 => x^2 - 8x + 8 = 0. Roots are 4 +/- 2*sqrt(2). Point (p,q) is (4+2*sqrt(2), 4+2*sqrt(2)). Tangent equation at (x0,y0) is (x-4)(x0-4) + (y-4)(y0-4) = 16. Plugging in (p,q) gives (x-4)(2*sqrt(2)) + (y-4)(2*sqrt(2)) = 16 => x+y = 8+4*sqrt(2). Intercepts are 8+4*sqrt(2). Area = 1/2 * (8+4*sqrt(2))^2 = 16(3+2*sqrt(2)).