SOLUTION: Find the constant value of k so that the line having equation
y =-3/4 x+k is tangent to the circle whose equation is (x-3)^2+(y+4)^2 = 25.
Algebra.Com
Question 207882: Find the constant value of k so that the line having equation
y =-3/4 x+k is tangent to the circle whose equation is (x-3)^2+(y+4)^2 = 25.
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
-----
your answer is k = 4.5
-----
derivation of your answer is shown below:
-----
the equation of the line tangent to the circle is y = -(3/4)*x + k
we'll call that line P
-----
this line will have to be perpendicular to the radius of the circle at that point.
we'll call that line R.
-----
the slope of the line R must be equal to the negative reciprocal of the slope of the line P because the radius of the circle and the line tangent to it are perpendicular to each other at that point.
-----
the slope of line P is -(3/4).
-----
the slope of line R must be (4/3) which is the negative reciprocal of (-3/4)
-----
we know that the radius must go through the center of the circle. This means that line R must go through the center of the circle.
-----
we know that the center of the circle is (3,-4) because the formula of the circle tells us so.
general formula of a circle is (x-h)^2 + (y-k)^2 = r^2 where (h,k) are the coordinates of the center of the circle.
since -h = -3, then h = 3
since -k = 4, then k = -4
center of the circle is (3,-4)
-----
general form of the equation of line R is y = (4/3)*x + b
where (4/3) is the slope and b is the y intercept.
to find the y intercept of that line, we plug in a value for one of the coordinates of that line and solve for b.
the center of the circle is one such point.
general form of line R becomes:
-4 = (4/3)*3 + b after we replace y with -4 and x with 3
-----
solving for b, we get b = -8
-----
equation of line R is y = (4/3)*x - 8
-----
we need to find the intersection of line R with the circle because that is also the point where line R will intersect with line P.
once we find that point, we can then solve the equation for line P which will result in allowing us to find k.
-----
we know the equation for line R.
we know the equation for the circle.
we can substitute the y value in the equation of the circle by the y value in the equation of line R. this will allow us to find the x value of the intersection of the line R with the equation of the circle.
once we find x, we can solve for y.
-----
here's how it was done.
-----
the equation of the circle is (x-3)^2 + (y+4)^2 = 25
the equation of line R is y = (4/3)*x - 8
-----
since (x-3)^2 equals x^2 - 6x + 9
and since (y+4)^2 equals y^2 + 8y + 16
then the equation of the circle expands to:
x^2 - 6x + 9 + y^2 +8y + 16 = 25
-----
since we know that y = (4/3)*x - 8, we can substitute for y in this equation.
-----
8*y becomes 8*(4/3)*x - 64
-----
y^2 becomes ((4/3)*x -8) * ((4/3)*x - 8) which becomes:
(16/9)*x^2 - 8*(4/3)*x - 8*(4/3)*x + 64 which becomes:
(16/9)*x^2 - 16*(4/3)*x + 64
-----
y^2 + 8y becomes:
(16/9)*x^2 - 16*(4/3)*x + 64 + 8*(4/3)*x - 64 which becomes:
(16/9)*x^2 - 8*(4/3)*x
-----
the equation of the circle becomes:
x^2 - 6x + 9 + (16/9)*x^2 - 8*(4/3)*x + 16 = 25
-----
in order to remove the denominators, we multiply both sides of this equation by 9 to get:
9x^2 - 54x + 81 + 16x^2 - 96x + 144 = 225
-----
combining like terms we get:
25x^2 - 150x + 225 = 225
-----
subtracting 225 from both sides of this equation gets:
25x^2 - 150x = 0
-----
dividing both sides of this equation by 25 gets:
x^2 - 6x = 0
-----
factoring out the x gets:
x * (x-6) = 0
which results in:
x = 0
or:
x = 6
-----
as shown on the graph, both these values for x are good.
-----
solving for y using line R and the value of x = 6 gets:
y = (4/3)*6 - 8
which results in:
y = 0
-----
solving for y using the original equation of the circle and the value of x = 6 gets:
(6-3)^2 + (y+4)^2 = 25 which becomes:
9 + y^2 + 8y + 16 = 25 which becomes:
y^2 + 8y = 0 which becomes:
y * (y-8) = 0 which becomes:
y = 0
or
y = -8
-----
when x = 6, y = 0 is good.
when x = 0, y = -8 is good.
this can be seen from the graph.
-----
we now have the intersection of line R with the equation of the circle.
we also have the intersection of line R with line P since both line R and line P intersect with the equation of the circle at that point.
-----
now we have a point for line P that we can use to solve that equation.
-----
the equation for line P is y = (-3/4)*x + k
if we plug in the value of 6 for x and the value of 0 for y, the equation becomes:
0 = (-3/4)*(6) + k
this results in:
k = (3/4)*(6) which results in:
k = 4.5
-----
RELATED QUESTIONS
The equation of the line tangent to the curve y=kx+8/k+x at x=-2 is y=x+4. What is the... (answered by richwmiller,jim_thompson5910)
Let k be a positive real number. The line x + y = k/2 and the circle x^2 + y^2 = 3x - 6y... (answered by CPhill)
find a value for k so that the line through (6,-3) and (k,-4) is parallel to the line... (answered by oberobic)
For what values of 'k' is the line y=x tangent to the curve with equation y=k/(x-1)... (answered by robertb)
The equation is y=(x+2)^2+k where k is a constant.
a)Find k when its tangent at x=1... (answered by ewatrrr)
find the value of k so that x^2+y^2+4x-6y+k=0 is the equation of circle of radius... (answered by josgarithmetic)
Find the value of k so that the line (2+k)x + (2-k)y=15 has slope... (answered by Nate)
Find the equation of the circle that passes through the point (-3,-4) and touches line... (answered by josgarithmetic)
find the value of k so that the equation 5x^2-8x+k=0 will have roots whose product is... (answered by stanbon)