What is the shortest distance between two circles, the first having centre A(5,3) and radius 12, and the other

Question

with centre B (2,-1) and radius 6?

Please Show your work.

2) If x and y are real numbers, determine all solutions (x,y) of the system of equations below.

x^2 -xy+8=0
x^2-8x+y=0

Please show your work.

3) Two parallel lines intersect the x-axis cutting off a line segment of length 3. The same lines also cut off a line segment along the y-axis of length 4. What is the perpendicular distance between these parallel lines?

a) 9/5 b)2 c) 12/5 d) 3 e) 16/5

Please show your work.

Thanks

Answer 1

What is the shortest distance between two circles, the first having centre A(5,3) and radius 12, and the other
with centre B (2,-1) and radius 6?

Please Show your work.

Length AB = sqrt( (5 - 2)² + (3 - -1)² ) = 5

So circle centre B is wholly contained inside circle centre A (sketch to see what I mean, if need be)

Draw line through A,B

So circle centre A radius 12 cuts line either 7 units from B on other side of B away from A (ie in A→B direction) or 19 units from B on same side as A (ie in B→A direction)
Also circle centre B radius 6 cuts line either 6 units from B on other side of B away from A (ie in A→B direction) or 6 units from B on same side as A (ie in B→A direction) Thus the closest the two circles get is 1 unit on the line on the opposite side to B to the side that A is.

2) If x and y are real numbers, determine all solutions (x,y) of the system of equations below.

x² - xy + 8 = 0
x² - 8x + y = 0

Please show your work

x² - xy + 8 = 0 ...... (1)
x² - 8x + y = 0 ........(2)
So y = 8x - x² .........(2a)

Substitute equation (2a) into equation (1)

x² - x(8x - x²) + 8 = 0

ie x² - 8x² + x³ + 8 = 0

ie x³ - 7x² + 8 = 0

When x = -1, x³ - 7x² + 8 = -1 - 7 + 8 = 0

Therefore (x + 1) is a factor (By factor theorem)

So x³ - 7x² + 8 = (x + 1)(x² - 8x + 8) = 0

So x + 1 = 0 or x² - 8x + 8 = 0

Thus x = -1 or x = ½(8 ± √[(-8)² - 4(1)(8)])
ie x = -1, 4 ± √8
= -1, 4 ± 2√2

When x = -1
y = -9

When x = 4 ± 2√2,
y = 8(4 ± 2√2) - (4 ± 2√2)²
= 32 ± 16√2 - (24 ± 16√2)
= 8

So the solutions are
x = -1 y = -9,
x = 4 + 2√2, y = 8,
x = 4 - 2√2, y = 8

3) Two parallel lines intersect the x-axis cutting off a line segment of length 3. The same lines also cut off a line segment along the y-axis of length 4. What is the perpendicular distance between these parallel lines?

a) 9/5 b)2 c) 12/5 d) 3 e) 16/5

Sketch

......y.|^
.........|./.../
........./.../
......../|4/
....../..|/
...../. /|
---/3/-|----->x

This gives the following geometry


.../ |
5/\ |4 (Sketch poor but as good as Y!A platform allows)
./_\|
. 3

Whence 3/5 = d/4

So d = 12/5 ← option c)

Answer 2

(1) The answer is 0, because the two circles intersect. The centers of the circles are separated by sqrt((5-2)^2+(3+1)^2) = 5, so each circle encloses the other circle's center, so the circles must intersect.

(2) x^2 - 8x + y = 0 ==> y = 8x - x^2 ==> x^2 - x(8x - x^2) + 8 = 0 ==> x^2 - 8x^2 + x^3 + 8 = 0 ==> x^3 - 7x^2 + 8 = 0 ==> (x+1)(x^2 - 8x + 8) = 0 ==> x=-1 or (x^2 - 8x + 8) = 0. Solving this last equation, x = (8 +/- sqrt(8^2 - 4*1*8)) / 2 = 4 +/- sqrt(32)/2 = 4 +/- 2*sqrt(2).

Therefore, the three solutions are: x=-1, x=4+2sqrt(2), x=4-sqrt(2).

(3) Without loss of generality, suppose that one of the lines crosses both the x and y axes at the origin. Then, the other line must cross the y-axis at y=4 and the x-axis at x=3. The axes and the second line segment (call it B) therefore form a right triangle whose hypotenuse is length sqrt(3^2+4^2) = 5. Let a be the angle formed by the x-axis and the segment B. sin a = 4/5. Define a line segment between the origin and segment B which is perpendicular to B. Call this segment C. A right triangle is then formed by the x-axis, C, and the portion of B between C and the x-axis, with the x-axis as the hypotenuse. Therefore, the length of C must be 3 * sin a = 12/5. The length of C is the distance between the two original lines.
==> (c) 12/5

Answer 3

1) What is the shortest distance between two circles, the first having centre A(5,3) and radius 12, and the other
with centre B (2,-1) and radius 6?

d = Distance between A and B is
R = radius of big circle = 12
r = radius of little circle = 6

d = √{(5-2)² + (3+1)²} = √(3² + 4²) = √(9 + 16) = √25 = 5

The distance between the centers of the circles is 5. And since the difference in radii is:

R - r = 12 - 6 = 6 > 5

The larger circle entirely contains the smaller circle. So the closest distance between two points on the circumference of the circles is:

R - (d + r) = 12 - (5 + 6) = 12 - 11 = 1

3) Two parallel lines intersect the x-axis cutting off a line segment of length 3. The same lines also cut off a line segment along the y-axis of length 4. What is the perpendicular distance between these parallel lines?

If we drop a perpendicular from the right angle to the hypotenuse we get to triangles, both congruent to the original triangle. Choosing one of them with a hypotenuse of 4 parallel to the y axis, the perpendicular distance is the shorter leg and its length is:

4(3/5) = 12/5

Answer 4

2)
Since they're both equal to zero, set them equal to each other
x^2 -xy +8=x^2-8x+Y
-xy+8= -8x+Y
8+8X = y+xy
8(1+x)= y(1+X)
8 = y
Then you substitute back and solve the quadractic using the quadratic solver
So Y=8 and X= 6.8284.... and 1.1715...