Let AB be the chord $4x-3y+5=0$ of the circle $x^2+y^2-2x+4y-20=0$. If $C=(7, 1)$ then the area of triangle ABC is?

The given circle x squared plus y squared minus 2x plus 4y minus 20 equals 0 has center (1, negative 2) and radius 5. The perpendicular distance from this center to the chord 4x minus 3y plus 5 equals 0 is the absolute value of (4(1) minus 3(negative 2) plus 5) divided by sqrt(4 squared plus negative 3 squared), which equals 15 by 5 or 3. Using the Pythagorean theorem, half the chord length is sqrt(5 squared minus 3 squared), which is 4, making the total chord length AB equal to 8. The perpendicular distance from C (7, 1) to the line 4x minus 3y plus 5 equals 0 is the absolute value of (4(7) minus 3(1) plus 5) divided by 5, which equals 6. The area of triangle ABC is one half times base AB times height, so 0.5 multiplied by 8 multiplied by 6 equals 24 square units.

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