Question

Write in slope intercept form the equation of the line tangent to the circle (x - 2)^2 + (y - 2)= 13 through the point (-1,4).​

Answer 1

The equation of the tangent line to the circle (x - 2)^2 + (y - 2)^2 = 13 passing through the point (-1, 4) in slope-intercept form is y = (3/2)x + 11/2.

Step-by-step explanation:

To find the equation of the line tangent to the circle (x - 2)^2 + (y - 2)^2 = 13 that passes through the point (-1, 4), we need to follow a few steps:

1. Find the center of the circle and the radius. The center is at (2, 2) and the radius is the square root of 13.
2. Find the slope of the radius at the point of tangency. Since the tangent line is perpendicular to the radius at the point of tangency, the slope of the tangent line will be the negative reciprocal of the slope of the radius.
3. Calculate the slope of the radius using the point (-1, 4) and the center (2, 2). Slope m = (4 - 2) / (-1 - 2) = 2 / -3.
4. The slope of the tangent line is the negative reciprocal, which is 3/2.
5. Use point-slope form to write the equation of the tangent line with the slope 3/2 and passing through (-1, 4). The point-slope form is y - y1 = m(x - x1). Plugging the values, we get y - 4 = (3/2)(x + 1).
6. Simplify to the slope-intercept form, y = mx + b, by distributing and isolating y. This gives us the final equation y = (3/2)x + (3/2) + 4 = (3/2)x + 11/2.

Therefore, the equation of the tangent line in slope-intercept form is y = (3/2)x + 11/2.