For the circle x^2+y^2+6x-4y+3=0
how would i go about finding:
a) the center and the radius
b) the equation of the tangent line at the point (-2,5)
any help is mch appreciated. thanks!
(a) Rewrite the equation as
x^2 + 6x + 9 + y^2 -4y + 4 +3 -13 = 0
(x+3)^2 + (y-2)^2 = 10 = (sqrt 10)^2
The should tell you that the center is at (-3, 2) and the circle radius is sqrt 10.
(b) First cmpute the slope dy/dx at (-2, 5)using implicit differentiation.
2x + 2y y' + 6 -4 y'= 0
y'(4 - 2y) = 2x + 6
y' = 2/(4-2y) = -1/3
Finally, write down the equation of s straight line passing through (-2,5) with that slope.
Verify my numbers. I may have goofed somewhere along the line
To find the center and radius of the given circle equation, which is in the standard form (x - h)^2 + (y - k)^2 = r^2, we need to rewrite the given equation:
x^2 + y^2 + 6x - 4y + 3 = 0
Rearrange the equation:
x^2 + 6x + y^2 - 4y = -3
Complete the square for both x terms and y terms by adding and subtracting appropriate constants:
(x^2 + 6x + 9) + (y^2 - 4y + 4) = -3 + 9 + 4
(x + 3)^2 + (y - 2)^2 = 10
Now we have the equation in the standard form, (x - h)^2 + (y - k)^2 = r^2, where (h, k) is the center of the circle, and r is the radius.
a) Comparing the given equation to the standard form equation, we can see that the center of the circle is (-3, 2), and the radius is √10.
b) To find the equation of the tangent line at the point (-2, 5), we need to find the slope of the tangent line and then use the point-slope form of a line.
First, find the derivative of the circle equation by differentiating implicitly with respect to x:
2(x + 3) + 2(y - 2) * dy/dx = 0
To find the slope of the tangent line at (-2, 5), substitute the coordinates into the derivative equation:
2(-2 + 3) + 2(5 - 2) * dy/dx = 0
2 + 6 * dy/dx = 0
dy/dx = -1/3
Now, using the point-slope form of a line (y - y1) = m(x - x1), where (x1, y1) is a point on the line and m is the slope, we can substitute the values:
(y - 5) = (-1/3)(x + 2)
Simplifying, we have:
y - 5 = (-1/3)x - 2/3
y = (-1/3)x + 13/3
Therefore, the equation of the tangent line at (-2, 5) is y = (-1/3)x + 13/3.