write the equation for a circle whose center is 5,3 and that is tangent to the line 3x+4y=12?

Question

thnk you

Answer 1

To find the point of tangency, you must take a line from the center that is perpendicular to the given line. To do this, you must find the slope of the given equation. Do this by putting it into y=mx+b form:

3x+4y=12
4y=12-3x
y=(-3/4)x+3

The slope of the given equation is -3/4. To find the slope of its perpendicular, take the negative reciprocal of the given slope, which is 4/3. Now, plug in the coordinates of the center to get the equation of the perpendicular:

y=(4/3)x+b
3=(4/3)*5+b
b=-11/3

So the equation of the perpendicular is y=(4/3)x-11/3, or 4x-3y=11.

The point of tangency is where these two lines intersect:

3x+4y=12
4x-3y=11

9x+12y=36
16x-12y=44
---------------
25x=80
x=16/5
48/5+4y=12
y=3/5

The radius of the circle is the distance from the center to the point of tangency:

r=sqrt((5-16/5)^2+(3-3/5)^2)
=sqrt(81/25+144/25)
=sqrt(9)=3

Finally, to write the equation of the circle, plug the center coordinates and radius into (x-h)^2+(y-k)^2=r^2
(x-5)^2+(y-3)^2=9

Answer 2

In order for the line 3x + 4y = 12 to be tangent to the circle whose center is (5,3), they must cross at exactly _one_ point.

The general equation for a circle whose center is at (5,3) is

(x - 5)^2 + (y - 3)^2 = r^2

However, since 3x + 4y = 12, then
4y = -3x + 12, and
y = (-3/4)x + 3, so if we plug this value in for y, we get

(x - 5)^2 + ( (-3/4)x + 3 - 3 )^2 = r^2
(x - 5)^2 + ( (-3/4)x )^2 = r^2
(x - 5)^2 + (9/16)x^2 = r^2
16(x - 5)^2 + 9x^2 = 16r^2

16(x^2 - 10x + 25) + 9x^2 = 16r^2
16x^2 - 160x + 400 + 9x^2 = 16r^2
25x^2 - 160x + 400 = 16r^2

25x^2 - 160x + 400 - 16r^2 = 0

At this point, it's worth nothing that we WANT exactly one value when solving for x. This is determined by the discriminant, which is b^2 - 4ac. We want b^2 - 4ac to equal 0. In this case
b = -160, a = 25, c = (400 - 16r^2), so

b^2 - 4ac = (-160)^2 - 4(25)(400 - 16r^2)
= 25600 - 40000 + 1600r^2
= -14400 + 16r^2

and we WANT this to equal 0, so

-14400 + 16r^2 = 0
16r^2 = 14400
r^2 = 900, therefore

Remember that the equation of our circle is

(x - 5)^2 + (y - 3)^2 = r^2
And now we know the value of r^2 (it's equal to 900), so

(x - 5)^2 + (y - 3)^2 = 900

Let's make sure they intersect at one point:

y = (-3/4)x + 3. Plugging this into our proposed circle equation:

(x - 5)^2 + ( (-3/4)x + 3 - 3)^2 = 900
(x - 5)^2 + ( (-3/4)x )^2 = 900
(x - 5)^2 + (9/16)x^2 = 900
16(x - 5)^2 + 9x^2 = 14400
16(x^2 - 10x + 25) + 9x^2 = 14400
16x^2 - 160x + 400 + 9x^2 = 14400
25x^2 - 160x + 400 - 14400 = 0
25x^2 - 160x - 14000 = 0
5x^2 - 32x - 2800 = 0

I apparently made an algebraic error somewhere.

Answer 3

The line above can be put into slope intercept form: y=-3/4x+3. Since the slope of this line is -3/4, the slope of the line perpendicular to this is 4/3. Using the point slope formula, the equation of the line passing through (5,3) having slope 4/3 is: y-3=4/3(x-5)...y=4/3x-11/3. Next, determine where this line and your oringinal line intersect by setting them equal to each other: -3/4x+3=4/3x-11/3.....7/12x=20/3....x=80/7. To determine the y coordinate of the intersection, substitute the value we just determined for x: y=4/3*80/7-11/3....y=243/21. Finally, calculate the distance from (5,3) to (11.42,11.57):sqrt(6.42^2+8.57^2)
=10.7. Therefore, the equation is: (x-5)^2+(y-3)^2=10.7^2

Answer 4

Step one million: discover the perpendicular line to y = 0.5x + 3 passing via the factor (5,-2) The perpendicular line has gradient -2 Now substitute into y -y1 = m(x-x1) y + 2 = -2( x -5) y +2 = -2x +10 y = -2x +8 Step 2: discover the intersection of those 2 strains y = -2x + 8 (one million) y = 0.5x +3 (2) (one million) = (2) 0.5x + 3 = -2x +8 2.5x = 5 x = 5/ 2.5 = 2 y = -2(2) +8 = 4 So ( 2,4) is the factor of intersection. Step3; discover the gap from (5,-2) to ( 2,4) d^2 = (5-2)^2 + (-2-4)^2 = 9 +36 = 40 5 = r^2 Step 4 Write down equation of the circle (x -5)^2 + (y+2)^2 = 40 5

Answer 5

center= 5,3 = h,k
the circle eqn is in form, (x-h)^2+(y-k)^2=r^2
so,the tangent is 3x+4y=12
from centre(5,3) the length of tangent is √S11 = √ 3(5)+4(3)-12
= √15
this is radius..
so,the eqn of circle is (x-5)^2 +(y-3)^2=15

hope this helps..