Final answer:
The equation of line AB tangent to the circle OP can be derived by finding the slope of the radius at the tangency point (5, 1) and then using the negative reciprocal for the slope of AB. None of the provided answer choices match the correct equation of the tangent line in slope-intercept form derived from the given information.
Step-by-step explanation:
The student is asked to find the equation of the line AB that is tangent to the circle OP at the point (5, 1). The given equation for OP is a circle equation in the general form, which can be rewritten to find the center and radius. Once the center is found, the radius to the tangent point (5, 1) can be used to determine the slope of the tangent line AB, since the radius of a circle is perpendicular to the tangent at the point of tangency. The equation of the line can then be found using the point-slope form and converted into slope-intercept form.
First, rewrite the circle equation x^2 + y^2 – 2x + 4y = 20 to find the center by completing the square:
- (x^2 - 2x + 1) + (y^2 + 4y + 4) = 20 + 1 + 4
- (x - 1)^2 + (y + 2)^2 = 25
This shows the circle has a center at (1, -2) and a radius of 5. Now use the gradient of OP's radius to (5, 1) to find the slope of the tangent line AB. Since the radius is perpendicular to AB, the slope of AB is the negative reciprocal of the slope of the radius.
The slope of the radius is (1 - (-2)) / (5 - 1) = 3/4, so the slope of the tangent line AB is -4/3. Using the point (5, 1), the point-slope form is y - 1 = (-4/3)(x - 5). Simplifying to slope-intercept form gives:
- y = (-4/3)x + (20/3) + 1
- y = (-4/3)x + (23/3)
None of the answer choices match this result. Please ensure there are no typos and that the question is transcribed correctly. If this equation is properly derived from the given information, then none of the provided answer choices (A-D) are correct.