Question

The equation of the tangent to the circle x2+y2-3x+4y=15 at the point (-1,2)

Answer 1

[AI-generated answer — Bot GPT 3.5]

To find the equation of the tangent line to the circle at a given point, we need to find the slope of the tangent line at that point and then use the point-slope form of a line.

Given the equation of the circle: x^2 + y^2 - 3x + 4y = 15

To find the slope of the tangent line, we need to find the derivative of the equation with respect to x:

Differentiating both sides of the equation:
d/dx (x^2 + y^2 - 3x + 4y) = d/dx (15)
2x + 2y * dy/dx - 3 + 4 * dy/dx = 0

We know that the point (-1, 2) is on the tangent line, so we substitute these values into the equation:
2(-1) + 2(2) * dy/dx - 3 + 4 * dy/dx = 0
-2 + 4 * dy/dx - 3 + 4 * dy/dx = 0
8 * dy/dx = 5
dy/dx = 5/8

So the slope of the tangent line is 5/8.

Now we can use the point-slope form of a line to find the equation of the tangent line:
y - y1 = m(x - x1)
y - 2 = (5/8)(x - (-1))
y - 2 = (5/8)(x + 1)
y - 2 = (5/8)x + 5/8
y = (5/8)x + 21/8

Therefore, the equation of the tangent line to the circle x^2 + y^2 - 3x + 4y = 15 at the point (-1,2) is y = (5/8)x + 21/8.