A circle is tangent to the line 3x-4y=32 and the center is at (0,7). Write the equation of the circle.?

Question

Umm, Help?

Answer 1

Nasty problem!

Ok, here is the game plan:

(1) Find equation of the radius that passes through tangent. (Technically, the equation of the line passing through the radius.)
(2) Find where radius meets the tangent, i.e. the point of tangency.
(3) Find distance from center to tangency point (i.e. the radius of the circle)
(4) Plug in center and radius into standard (x-x1)^2+(y-y1)^2=r^2 to get equation for circle. Here, (x1,y1) is the center, and r is the radius.

Starting with(1), first re-write the equation in form y=mx-b. We have

3x-4y=32

which becomes

y=3/4x-8.

This has slope 3/4. The slope of the radius must be perpendicular to the tangent, so it has slope -4/3. (The slope of a line perpendicular to a 2nd line is -1/m, where m is the slope of the 2nd line.)

Using point slope form, we have

y-7=(-4/3)x

(Point-slope form is y-y1=m(x-x1). This is the equation for a line passing through (x1,y1) with slope m.)

(2) To find where the radius and tangency meet, solve th simultaneous set of equations

y-7=(-4/3)x (radius)

and

y=(3/4)x-8 (tangent)

Substituting the 2nd equation into the first one gives

(-4/3)x=(3/4)x-8-7

which gives

x=36/5=7.2,

and plugging this into the 2nd equation gives

y=(3/4)(7.2)-8=-2.6

So the point of tangency is (7.2,-2.6)

(3) Now, the distance from the center of the circle to this point is the square root of the squared differences of the coordinates of the points (i.e. Pythagorean theorem):

sqrt( (7.2-0)^2+(-2.6-7)^2)=12

(4) So the equation is just:

x^2+(y-7)^2=144

Good luck, hope this helped! I tried to avoid using calculus or obscure formulas.

Answer 2

Find the equation of the circle tangent to the line 3x - 4y = 32 with center (0,7).

Rewrite the equation of the line and set it equal to zero.

3x - 4y = 32
3x - 4y - 32 = 0

Find the distance d from the center of the circle to the given line using the distance formula. The formula for the distance from a point (h,k) to a line with equation ax + by + c = 0 is:

d = | ah + bk + c | / √(a² + b²)

Apply this formula to the problem at hand.

d = | 3*0 - 4*7 - 32 | / √[3² + (-4)²] = | 0 - 28 - 32 | √(9 + 16)

d = 60 / √25 = 60 / 5 = 12

The formula for a circle with center (h,k) and radius r is:

(x - h)² + (y - k)² = r²

Apply this formula to the problem at hand.

(x - 0)² + (y - 7)² = 12²

x² + (y - 7)² = 144

Answer 3

Ok, let's do it :

we have the center :

x^2 + (y - 7)^2 = R^2

tangent line : 3/4*x - 8 = y

slope of the line = 3/4 = dy / dx

using implicit differentiation :

2x + 2(y-7)*dy/dx = 0

dy/dx = 3/4 >>> that's the slope for a certain point of the circle

x + (y-7)*3/4 = 0

-4x = 3y - 21 >>> one equation

21 = 4x + 3y >>> equation of the line

Now we can get values for x and y, and replace them to find R

32 = 3x - 4y

84 = 16x + 12y

96 = 9x - 12y

180 = 25x

x = 7.2 and y = -2.75

replacing on :

x^2 + (y - 7)^2 = R^2

R = 12.12

(x - 7.2)^2 + (y - 7)^2 = 12.12^2 = 146.9

Hope that helps

Answer 4

the eq. of radius will be
first find the m for the tangent:
y=3/4x-8
m=3/4
as radius and tangent are perpendicular then
m2=-4/3
y=-4x/3+c
now to find c plug in the centre of circle
7=0+c
then 7=c
3y=-4x+21
now find what is the point where both tangent and radius meet
solve both eq. simultaneously
4(4x+3y=21)
3(3x-4y=31)

16x+12y=84
9x-12y=93
25x=177
x=177/25
3y=21-4(177/25)
y=-2.44
(7.08,-2.44)
now eq. of circle will be
(x-h)^2+(y-k)^2=r^2
(7.08)^2+(-2.44-7)^2=r^2
r^2=139.24
eq. is
x^2+(y-7)^2=139.24 ans

Answer 5

is this circles day today or what?