How do you write an equation of the line tangent to #(x-1)^2+(y-1)^2=25# at the point (4,-3)?

Question

Luke Alvarez · Accepted Answer

#y=3/4x-6# The equation is that of a circle centred at point #O(1,1)#, Let #P# be the point #(4,-3)#. The tangent at #P# is at right angles to the radius at #P#. But the gradient of radius #OP# is #(-3-1)/(4-1)=-4/3#. Therefore the gradient of the tangent is #(-1)/(-4/3)=+3/4#. Therefore the equation of the tangent is #y=3/4 x + c# for some #c#. Since #P# lies on the tangent, #-3=3/4 4+c#. Hence #c=-6#.

Emilia Aguilar · Answer

#y = 3/4x - 6#

Expand the equation of the circle.

#x^2 - 2x + 1 + y^2 - 2y + 1 = 25#

#x^2 + y^2 - 2x - 2y + 2 = 25#

Differentiate both sides with respect to x using implicit differentiation and the power rule.

#d/dx(x^2 + y^2 - 2x - 2y + 2) = d/dx(25)#

#2x + 2y(dy/dx) - 2 - 2(dy/dx) = 0#

#2y(dy/dx) - 2(dy/dx) = 2 - 2x#

#dy/dx(2y - 2) = 2 - 2x#

#dy/dx = (2 - 2x)/(2y - 2)#

#dy/dx = (2(1 - x))/(2(y - 1))#

#dy/dx = (1 -x)/(y - 1)#

#dy/dx = -(x - 1)/(y - 1)#

The slope of the tangent is given by evaluating #f(x, y)# inside the derivative.

#m_"tangent" = -(4 - 1)/(-3 - 1)#

#m_"tangent" = -3/(-4)#

#m_"tangent" = 3/4#

The equation of the tangent is therefore:

#y - y_1 = m(x -x_1)#

#y - (-3) = 3/4(x - 4)#

#y + 3 = 3/4x - 3#

#y = 3/4x - 6#

Hopefully this helps!

Ezekiel Alexander · Answer

undefined To write the equation of the line tangent to the circle (x-1)^2 + (y-1)^2 = 25 at the point (4,-3), we need to find the slope of the tangent line and the coordinates of the point of tangency.

First, we differentiate the equation of the circle with respect to x to find the derivative of y with respect to x.

2(x-1) + 2(y-1) * dy/dx = 0

Simplifying the equation, we get:

dy/dx = - (x-1) / (y-1)

Next, we substitute the coordinates of the point (4,-3) into the derivative equation to find the slope of the tangent line at that point.

dy/dx = - (4-1) / (-3-1) = -3/4

Now, we have the slope of the tangent line.

Using the point-slope form of a linear equation, we can write the equation of the tangent line:

y - y1 = m(x - x1)

Substituting the values of the slope (-3/4) and the point of tangency (4,-3) into the equation, we get:

y - (-3) = (-3/4)(x - 4)

Simplifying the equation, we have:

y + 3 = (-3/4)x + 3

Rearranging the equation, we get the final equation of the line tangent to the circle at the point (4,-3):

y = (-3/4)x + 3 - 3

Simplifying further, we have:

y = (-3/4)x