Analytic Goemetry And Calculus :MAT 102
1. Find the equation of the circle passing through the points P(2, 1), Q (0, 5).
and R (-1, 2).
A. x²+ y²+6x-2y - 5 =0
B. X² + y²- 6x- 2y +5= 0
C. x²+ y²+ 6x + 2y -5 = 0
D. x² + y²- 2x-6y + 5 = 0
2. Given that y + 2x - 4 = 0 intersects the circle (x+2)²+(y-4)²=5 find the point(s) of the intersection of the circle and
the line.
A. (1/5, 18/5) and (1,6)
B. (1/5, 18/5) and (-1,6)
C.(-1, 18/5) and (1/5, 6)
D. (1/5,-1) and (18/5, 6)
3. If the distance between the points (3, 9) and (8, x) is 13, find the values of x.
A.-2
B.-3
C.2
D.3
4. Compute the center and radius of the circle x²+y²+ 6x +8y +9=0.
A.(3, 4), r=4
B.(4,3), r=4
C(-4,3),r=4
D.(-3,-4), r=4
5. Find the equation of a line which passes through A (4,-1) and is parallel to
the y-axis.
A.Y=4
B.Y=-1
C.X=-1
D.X=4
Answers
1. To find the equation of the circle passing through points P(2, 1), Q(0, 5), and R(-1, 2), we can start by -- knowing that the general equation of a circle is given by (x-h)² + (y-k)² = r². Since it passes through these points, r will be equal to the distance of these points to the center (h, k). We want the equation in the form x² + y² + ax + by + c = 0, so let's solve using the given points:
Put the circle formula for each of the points:
P: (2-h)² + (1-k)² = r² (1)
Q: (0-h)² + (5-k)² = r² (2)
R: (-1-h)² + (2-k)² = r² (3)
Since all three equations have r², we can create the following systems of equations (1)-(2), (1)-(3), (2)-(3):
-2h + 2k = -1 (4)
4h+6k = 15 (5)
6h-2k = 20
From (4), we can express k in terms of h: k = h - 1/2
Substitute this expression in (5) to solve for h: h = 1
Now, put the h value in (4) to solve for k: k = 1/2
Finally, substitute h and k values in any of the circle formulas (e.g., 1): (1 + 1)² + (1 - 0.5)² = r²
r² = 5
So, the circle equation is (x - 1)² + (y - 1/2)² = 5. Expanding and organizing gives us x² + y² + 6x - 2y - 5 = 0. The correct option is A.
2. To find the intersection points of a circle and a line, we substitute the value of y from the line equation into the circle equation, and solve for x.
y = -2x + 4
Circle equation: (x + 2)² + (y - 4)² = 5
Substitute y:
(x + 2)² + (-2x + 4 - 4)² = 5
(x + 2)² + (-2x)² = 5
x² + 4x + 4 + 4x² = 5
5x² + 4x - 1 = 0
Using the quadratic formula to solve for x, we obtain x = 1/5 and x = -1.
Corresponding y values are y = 18/5 (-2x + 4 for x = 1/5) and y = 6 (for x = -1)
So, the points of intersection are (1/5, 18/5) and (-1, 6). The answer is B.
3. Using the distance formula, we can find the value of x for which the distance between (3, 9) and (8, x) is 13.
√((8 - 3)² + (x - 9)²) = 13
Solving for x²:
5² + (x - 9)² = 13²
(x - 9)² = 144 - 25
(x - 9)² = 119
Taking the square root of both sides and solving for x, we get two values: x = -3 and x = 3. The correct answer is D.
4. First, complete the square to find the forms (x-h)² and (y-k)²:
x² + 6x + (6/2)² + y² + 8y + (8/2)² = -9 + (6/2)² + (8/2)²
(x + 3)² + (y + 4)² = 4²
Thus, the center is (-3, -4), and the radius is 4. The correct answer is D.
5. For a line parallel to the y-axis, the x-coordinate remains constant. Since the line passes through point A(4, -1), the equation will be x = 4, which is answer D.
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