Give an equation of a circle with center at P(3,4) and the circle is tangent to the line y=3x+2?

Question

I have to find the equation and draw everything on the x,y axis.

Answer 1

The equation of a circle with center (h,k) and radius r is equal to

(x - h)^2 + (y - k)^2 = r^2

Since the center is at (3,4), we plug in 3 for h and 4 for k.

(x - 3)^2 + (y - 4)^2 = r^2

The circle is tangent to the line y = 3x + 2, so there can only be ONE solution when substituting y = 3x + 2 into the circle equation.

(x - 3)^2 + (3x + 2 - 4)^2 = r^2
(x - 3)^2 + (3x - 2)^2 = r^2
x^2 - 6x + 9 + 9x^2 - 12x + 4 = r^2
10x^2 - 18x + 13 = r^2
10x^2 - 18x + (13 - r^2) = 0

In order for this to have one solution, the discriminant (b^2 - 4ac) must equal 0. That is,

(-18)^2 - 4(10)(13 - r^2) = 0

We solve for this

324 - (520 + 40r^2) = 0
324 - 520 + 40r^2 = 0
-196 + 40r^2 = 0
40r^2 = 196
r^2 = 4.9
r = +/- sqrt(4.9)

That means the equation of our circle which is tangent to the line y=3x+2 is

y = (x - 3)^2 + (y - 4)^2 = 4.9

Answer 2

well draw the line first. it will intersect points (-1,-1), (0.2), and (1,5) and of coarse keep going on each end. P is (3,4) so plot that. fin the shortest distance from P to the line, that is the radius of the circle. to find the radius make a right triangle with the hypotenuse being the line from point (1,5) to P. the otehr sides will be 1 and 2. solve for the hypotenuse using the pathagorean theorum (a-squared + b-squared=c-squared) c being the hypotenuse (duh) and that will give u the radius of the circle. from ther u can use 2piR or piR-squared to find whatever u want to know about the circle

Answer 3

The distance R between P and any point on line is: R(x) = sqrt((y-Py)^2 + (x-Px)^2) = sqrt((y-4)^2 + (x-3)^2), the tangent circle must have minimal R when R’(x)=0;
R’(x) = {(y-4)*y’+(x-3)}/R(x) = 0, if (y-4)*y’+x-3=0;
As y=3x+2 and y’=3, then (3x+2–4)*3 + x-3 = 0, hence x=0.9 and y=3*0.9+2=4.7 and R=sqrt((4.7-4)^2 + (0.9-3)^2)=2.213594