find an equation of the line tangent to the circle (x-1)^2+(y-1)^2=25 at the point (4,-3)
Can please someone explain this to me and show all work very confused
Find the derivative implicityl
2(x-1) + 2(y-1) dy/dx = 0
plug in (4,-3) , find dy/dx, which is the slope
Now that you have the slope and a point, use the method you learned to find the equation of the straight line
To find the equation of the line tangent to a circle at a given point, we can use the properties of circles and slopes of tangent lines.
First, let's rewrite the equation of the circle in standard form:
(x - 1)^2 + (y - 1)^2 = 25
x^2 - 2x + 1 + y^2 - 2y + 1 = 25
x^2 + y^2 - 2x - 2y - 23 = 0
Next, we need to find the slope of the tangent line at the point (4, -3). The slope of a tangent line to a circle is perpendicular to the line connecting the center of the circle to the point of tangency.
The given circle has its center at (1, 1), so the slope of the line connecting (1, 1) to (4, -3) can be found using the formula:
m = (y2 - y1) / (x2 - x1)
= (-3 - 1) / (4 - 1)
= -4 / 3
Since the slope of the line connecting the center to the point of tangency is -4/3, the slope of the tangent line will be the negative reciprocal of this slope. In other words, if the center-to-point slope is m1, the slope of the tangent line, m2, will be given by m2 = -1 / m1.
m2 = -1 / (-4/3)
= 3/4
Now that we have the slope of the tangent line, we can use the point-slope form of the equation of a line to determine the equation. The point-slope form is given by:
y - y1 = m(x - x1)
With the point (4, -3) and the slope 3/4, we can substitute these values into the equation to get:
y - (-3) = (3/4)(x - 4)
y + 3 = (3/4)x - 3
y = (3/4)x - 6
Therefore, the equation of the line tangent to the circle (x - 1)^2 + (y - 1)^2 = 25 at the point (4, -3) is y = (3/4)x - 6.