a. the line y=7x cuts the circle x^2+y^2=50 at A and B. find the coordinates of A and B.

b. show that C(5;5) lies on the circle.

c. show that ACB angle C=90degrees. what can you deduce about AB?

d.find the equation of the tangent to the circle at C

nice problem:

b) the easiest, just sub in (5,5) to show it works

a) since y = 7x, sub the y into the circle equation
x^2 + (7x)^2 = 50
x^2 + 49x^2 = 50
x^1 = 1
x = ± 1

if x=1 , y = 7 , call that A(1,7)
if x = -1 , y = -7 , call that B(-1, -7)

c) slope AC = (5-7)/(5-1) = -2/4 = -1/2
slope BC = (5+7)/(5+1) = 12/6 = +2
since they are negative reciprocals of each other, they must form a 90° angle
AB must be a diameter of the circle.

d) slope of OC = (5-0)/(5-0) = 1
slope of the tangent must be -1
y-5 = -1(x-5)
y-5 = -x + 5
x+y = 10 is the tangent equation

thanks alot now I understand

a. To find the coordinates of points A and B where the line y = 7x intersects the circle x^2 + y^2 = 50, we need to solve the simultaneous equations formed by substituting y = 7x into the equation of the circle.

Substituting y = 7x into x^2 + y^2 = 50:

x^2 + (7x)^2 = 50
x^2 + 49x^2 = 50
50x^2 = 50
x^2 = 1
x = ±1

Substituting these values of x back into y = 7x to find y:

For x = 1:
y = 7 * 1 = 7
Coordinates of point A = (1, 7)

For x = -1:
y = 7 * -1 = -7
Coordinates of point B = (-1, -7)

Therefore, point A is (1, 7) and point B is (-1, -7).

b. To show that point C(5, 5) lies on the circle, we need to substitute the coordinates of C into the equation of the circle and check if the equation holds true.

Substituting x = 5 and y = 5 into x^2 + y^2 = 50:
(5)^2 + (5)^2 = 50
25 + 25 = 50
50 = 50

Since both sides of the equation are equal, it confirms that point C(5, 5) lies on the circle x^2 + y^2 = 50.

c. To show that angle C is 90 degrees (a right angle), we need to prove that vectors AB and BC are perpendicular to each other. If two vectors are perpendicular, then their dot product is zero.

The equation for the line passing through A and B, given by y = 7x, can be rewritten as 7x - y = 0.

For vector AB, the direction vector is (1 - (-1), 7 - (-7)) = (2, 14).
For vector BC, the direction vector is (5 - 1, 5 - 7) = (4, -2).

Now, calculate the dot product of AB and BC:

(2, 14) · (4, -2) = (2 * 4) + (14 * -2) = 8 - 28 = -20

Since the dot product is -20 (not equal to zero), it means that AB and BC are not perpendicular, and angle C is not 90 degrees.

Therefore, we can conclude that AB is not perpendicular to BC.

d. The equation of a tangent to a circle at a given point can be found using the point-slope form of a line equation. The point-slope form is given by y - y1 = m(x - x1), where (x1, y1) is the given point, and m is the slope of the tangent line.

The equation of the circle is x^2 + y^2 = 50, and the point of tangency is C(5, 5).

To find the slope of the tangent line at C, we first find the derivative of the circle equation with respect to x:

d/dx (x^2 + y^2) = d/dx (50)
2x + 2y * dy/dx = 0
dy/dx = -(2x) / (2y)

Substituting the coordinates of C(5, 5) into the derivative equation:

dy/dx = -(2 * 5) / (2 * 5)
dy/dx = -1

So, the slope of the tangent line at C is -1.

Now, using the point-slope form with the given point (5, 5) and slope -1:

y - 5 = -1(x - 5)
y - 5 = -x + 5
y = -x + 10

Therefore, the equation of the tangent to the circle at C is y = -x + 10.