a) Determine the equation of the circle center at the origin through A(-2;4).Draw a sketch
b)Determine the equation of the tangent to the circle at A.
c) This tangent cuts the x-axis at B. Determine the length of AB.
d) Determine the equation of the other tangent at the circle from B.
a) √-2^2+4^2
= √4+16
= √20
The general equation of a circle whose centre is (a,b) and radius r is:
(x-a)^2 + (y-b)2 = r2
Therefore, the equation is x2 + y2 = 20.
a) Equation is: x^2 + y^2 = 20
(a) as above, x^2+y^2 = 20
(b) dy/dx = -x/y
so (-2,4) the slope is 1/2, making the tangent line
y-4 = 1/2 (x+2)
(c) at B=(±√20,0) the length of AB is √((2±√20)^2+4^2)
(d) since y=0, the line is vertical: x=±√20
a) To determine the equation of a circle centered at the origin through point A(-2, 4), we can use the general equation of a circle: (x - h)^2 + (y - k)^2 = r^2, where (h, k) represents the coordinates of the center and r represents the radius.
Since the circle passes through the origin (0, 0), the center is at (0, 0). We can substitute these values into the general equation to get: (x - 0)^2 + (y - 0)^2 = r^2, which simplifies to x^2 + y^2 = r^2.
To find the value of r, we can substitute the coordinates of point A(-2, 4) into the equation: (-2)^2 + 4^2 = r^2. Simplifying this equation, we get 4 + 16 = r^2, giving us r^2 = 20.
Therefore, the equation of the circle centered at the origin through A(-2, 4) is x^2 + y^2 = 20.
To draw a sketch, you can plot the center of the circle at the origin (0, 0) and plot point A(-2, 4) on the graph. Then, draw the circle using the equation x^2 + y^2 = 20, ensuring that all points on the circle are equidistant from the center.
b) To determine the equation of the tangent to the circle at point A, we can use the fact that a tangent to a circle is perpendicular to the radius at the point of tangency.
Since the center of the circle is at the origin (0, 0), the radius of the circle coincides with point A(-2, 4). The slope of the radius can be found using the formula m = (y2 - y1) / (x2 - x1), where (x1, y1) and (x2, y2) are the coordinates of two points on the line.
In this case, the slope of the radius passing through the origin and point A is m = (4 - 0) / (-2 - 0) = -2.
Since the tangent is perpendicular to the radius, its slope will be the negative reciprocal of the radius's slope. Therefore, the slope of the tangent is 1/2.
Using the point-slope form of a line, y - y1 = m(x - x1), we can substitute the values y1 = 4, x1 = -2, and m = 1/2 into the equation to get y - 4 = (1/2)(x + 2).
Simplifying this equation, we get y = (1/2)x + 3, which is the equation of the tangent to the circle at point A(-2, 4).
c) To determine the length of AB, we need to find the x-coordinate of point B where the tangent intersects the x-axis.
To find this, we substitute y = 0 into the equation y = (1/2)x + 3, which gives us 0 = (1/2)x + 3. Solving for x, we get x = -6.
Therefore, point B is (-6, 0). To find the length of AB, we can use the distance formula: d = sqrt((x2 - x1)^2 + (y2 - y1)^2).
Plugging in the coordinates of A(-2, 4) and B(-6, 0), we get d = sqrt((-6 - (-2))^2 + (0 - 4)^2) = sqrt((-4)^2 + (-4)^2) = sqrt(16 + 16) = sqrt(32).
Hence, the length of AB is sqrt(32).
d) To determine the equation of the other tangent at the circle from point B(-6, 0), we can follow a similar procedure as in part b).
The slope of the radius passing through the origin (0, 0) and point B(-6, 0) can be found as m = (0 - 0) / (-6 - 0) = 0.
Since the tangent is perpendicular to the radius, its slope will be the negative reciprocal of the radius's slope. Therefore, the slope of the tangent is undefined.
Since the tangent is vertical and passes through point B(-6, 0), its equation will be of the form x = b, where b is the x-coordinate of point B.
Hence, the equation of the other tangent at the circle from B is x = -6.