Frazer cycles the first 20 miles at an average speed of 21 mph. The second part is more uphill, and he only manages 13mph. By wh
1 answer:
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Answer:
6. C: {x^2 +(y-1)^2 =2; x+y = 3}
15. C: The line does not intersect the circle.
Step-by-step explanation:
The formula for the distance (d) from a point (x, y) to a line ax+by=c is ...
d = |ax+by-c|/√(a^2+b^2)
The formula for a circle centered at (h, k) with radius r is ...
(x -h)^2 +(y -k)^2 = r^2
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6. Comparing the circle equation to the generic equation, we find (h, k) = (0, 1) and r = √2. Then we want to find the line that is distance √2 from the center of the circle. Our line equation is x+y=c for some value of c that we want to find.
d = √2 = |0 +1 -c|/√(1^2+1^2)
2 = |1-c|
±2 = 1-c
c = 1±2 = -1 or 3
The line that is tangent to the circle is the one of choice C: x+y = 3
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The attached graph shows the lines for all 4 answer choices. The point of tangency is (1, 2), so x+y=1+2=3.
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15. The circle is centered at (4, 1) and has radius 3. The distance from the circle center to the line is ...
d = |2(4) -(1)|/√(2^2+(-1)^2) = 7/√5 ≈ 3.13
The distance from the circle center to the line is more than the radius of the circle, so there can be no points of intersection.
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Alternate solution
You can substitute for y using the equation of the line. Then the circle equation becomes ...
(x -4)^2 + (2x -1)^2 = 9
x^2 -8x +16 +4x^2 -4x +1 = 9
5x^2 -12x +8 = 0
The discriminant of this quadratic is ...
b^2 -4ac = (-12)^2 -4(5)(8) = 144-160 = -16
Since this value is negative, there can be no real solutions, meaning the line does not intersect the circle.
