a)determine whether the point A(-2, -6) lies on the circle defined by x^2 + y^2 = 40
b) find an equation for the radius from the origin O to point A
c) Find an equation for the line that passes through A and is perpendicular to OA
a) When x = -2, in order to be on the circle, y must satisfy the equation
4 + y^2 = 40. That means
y^2 = 36.
y can be either 6 or -6. Therefore the point (-2,-6) IS on the circle.
b) The origin of the circle is (0,0). The equation of the circle tells you that. A line passing from there to the point (-2,-6) has slope (-6/-2) = 3. Its equation is y = 3x. (There is no constant term, since the y-intercept is 0)
c)The slope of a perpendicular line must be -1/3. It must pass through (-2,-6) go thorugh A. The equation of that line is:
[y-(-6)]/[x-(-2)] = -1/3
(y+6)/(x+2) = -1/3
y+6 = -x/3 -2/3
y = -x/3 -20/3
Omg thnx u sooooooo muchhhhhhhhhhhhhhhhhh
a) Well, A(-2, -6), did you pay rent to the circle x^2 + y^2 = 40? Let's see if you fit in. Plug in the values for x and y. We get (-2)^2 + (-6)^2 = 4 + 36 = 40. Voila! You're in! A is on the circle!
b) Ah, the equation for the radius from the origin O to point A. Let's calculate the distance. Remember the Pythagorean theorem? It's like the OG math equation! We have x^2 + y^2 = r^2, where (x, y) is point A and r is the radius. Plugging in the values, we get (-2)^2 + (-6)^2 = r^2, giving us 40 = r^2. Taking the square root of both sides, we find the radius is ±√40. But since we can't have a negative radius (imagine that), we can say the equation for the radius is r = √40.
c) Now, let's find an equation for the line that passes through A and is perpendicular to OA. So, the slope of any line perpendicular to OA is the negative reciprocal of the slope of OA. Since OA is a line connecting point A and the origin O, its slope is y2 - y1 / x2 - x1, which is -6 - 0 / -2 - 0 = 3. Now, the negative reciprocal of 3 is -1/3. We need a point on the line, and we already know that A(-2, -6) falls on it. So, using the point-slope form, y - y1 = m(x - x1), we can say y - (-6) = -1/3(x - (-2)), which simplifies to y + 6 = -1/3(x + 2). Tada! That's the equation for the line passing through A and perpendicular to OA.
a) To determine whether the point A(-2, -6) lies on the circle defined by x^2 + y^2 = 40, we substitute the x and y values of point A into the equation and check if the left-hand side equals the right-hand side.
Plugging in x = -2 and y = -6:
(-2)^2 + (-6)^2 = 4 + 36 = 40
The left-hand side equals the right-hand side, so the point A(-2, -6) lies on the circle defined by x^2 + y^2 = 40.
b) The radius from the origin O to point A can be found by using the distance formula, which is the square root of the sum of the squares of the differences in the x and y coordinates.
Let O(0, 0) be the origin.
Using the distance formula:
radius = √[(x2 - x1)^2 + (y2 - y1)^2]
radius = √[(0 - (-2))^2 + (0 - (-6))^2]
radius = √[(0 + 2)^2 + (0 + 6)^2]
radius = √[2^2 + 6^2]
radius = √[4 + 36]
radius = √40
radius = 2√10
So, the equation for the radius from the origin O to point A is r = 2√10.
c) To find an equation for the line that passes through point A and is perpendicular to OA, we first need to find the slope of OA.
The slope of OA is given by the formula:
slope = (y2 - y1) / (x2 - x1)
Let O(0, 0) be the origin and A(-2, -6).
Substituting the values:
slope = (-6 - 0) / (-2 - 0)
slope = -6 / -2
slope = 3
Since the line perpendicular to OA will have a slope that is the negative reciprocal of 3 (which is -1/3), we can use the point-slope form of a line to write the equation.
Using point-slope form:
y - y1 = m(x - x1)
Where m is the slope and (x1, y1) is a point on the line (in this case, A).
Using A(-2, -6):
y - (-6) = (-1/3)(x - (-2))
y + 6 = (-1/3)(x + 2)
Expanding and simplifying:
y + 6 = (-1/3)x - 2/3
Rearranging the equation to slope-intercept form:
y = (-1/3)x - 2/3 - 6
y = (-1/3)x - 20/3
So, the equation for the line that passes through A and is perpendicular to OA is y = (-1/3)x - 20/3.
To determine whether the point A(-2, -6) lies on the circle defined by x^2 + y^2 = 40, we can substitute the x and y values of the point into the equation and check if the equation holds true.
a) Substituting the x and y values of A into the equation x^2 + y^2 = 40, we have:
(-2)^2 + (-6)^2 = 4 + 36 = 40
Since this equation holds true, we can conclude that the point A(-2, -6) lies on the circle defined by x^2 + y^2 = 40.
b) To find the equation for the radius from the origin O to point A, we can use the distance formula. The origin is the point (0, 0), so the distance formula is:
Distance = sqrt((x2 - x1)^2 + (y2 - y1)^2)
Substituting the coordinates of O(0, 0) and A(-2, -6) into the formula, we have:
Distance = sqrt((-2 - 0)^2 + (-6 - 0)^2)
Distance = sqrt((-2)^2 + (-6)^2)
Distance = sqrt(4 + 36)
Distance = sqrt(40)
Therefore, the equation for the radius from the origin O to point A is:
Distance = sqrt(40)
c) To find an equation for the line that passes through A(-2, -6) and is perpendicular to OA, we need to find the slope of OA first. The slope of a line passing through two points (x1, y1) and (x2, y2) is given by:
Slope = (y2 - y1) / (x2 - x1)
Substituting the coordinates of O(0, 0) and A(-2, -6) into the slope formula, we have:
Slope = (-6 - 0) / (-2 - 0)
Slope = -6 / -2
Slope = 3
The line perpendicular to OA will have a slope that is the negative reciprocal of 3. Let's call this slope m.
m = -1/3
Now we have the slope and a point (A) on the line. We can use the point-slope form of a line, which is:
y - y1 = m(x - x1)
Substituting the values of m = -1/3 and the coordinates of A(-2, -6) into the equation, we have:
y - (-6) = (-1/3)(x - (-2))
y + 6 = (-1/3)(x + 2)
Simplifying the equation, we get:
y + 6 = (-1/3)x - 2/3
y = (-1/3)x - 2/3 - 6
y = (-1/3)x - 20/3
Therefore, the equation for the line that passes through A(-2, -6) and is perpendicular to OA is:
y = (-1/3)x - 20/3