Find the equation of the diameter of the circle x^2+y^2-6x+2y=15, which, when produced, passes through the point (8,-2).
completing the square will give us the centre ....
x^2 - 6x + .... + y^2 + 2y + ... = 15
x^2 - 6x + 9 + y^2 + 2y + 1 = 15 + 9+1
(x-3)^2 + (y+1)^2 = 25
centre is (3,-1)
so you want the line through (3,-1) and (8,-2)
slope = (-2+1)/(8-3) = -1/5
y+1 = (-1/5(x-3)
5y + 5 = -x + 3
x + 5y = -2 or y = (-1/5)x - 2/5
thanks
To find the equation of the diameter of the circle, we need to find the center of the circle and another point on the circle.
Step 1: Rewrite the equation of the circle in standard form by completing the square.
x^2 - 6x + y^2 + 2y = 15
(x^2 - 6x + 9) + (y^2 + 2y + 1) = 15 + 9 + 1
(x - 3)^2 + (y + 1)^2 = 25
Step 2: Comparing the equation to the standard form of a circle (x - h)^2 + (y - k)^2 = r^2, we can see that the center of the circle is at (h, k) = (3, -1) and the radius is r = 5.
Step 3: Now, we need to find the equation of the line passing through the point (8, -2) and the center of the circle (3, -1). Since the line passes through two points, we can use the point-slope form.
m = (y2 - y1) / (x2 - x1)
m = (-2 - (-1)) / (8 - 3)
m = -1 / 5
Using the point-slope form, we have:
y - y1 = m(x - x1)
y - (-1) = (-1/5)(x - 3)
y + 1 = (-1/5)(x - 3)
5(y + 1) = -(x - 3)
5y + 5 = -x + 3
x + 5y - 2 = 0
Therefore, the equation of the diameter of the circle is x + 5y - 2 = 0.
To find the equation of the diameter of a circle given its equation and a point it passes through, follow these steps:
1. Write the equation of the circle in the standard form by completing the square for both x and y terms.
The given equation is x^2 + y^2 - 6x + 2y = 15. Rearranging the terms, we have:
(x^2 - 6x) + (y^2 + 2y) = 15
To complete the square for x, add the square of half the coefficient of x (which is (-6/2)^2 = 9) to both sides of the equation:
(x^2 - 6x + 9) + (y^2 + 2y) = 15 + 9
(x - 3)^2 + (y^2 + 2y) = 24
To complete the square for y, add the square of half the coefficient of y (which is (2/2)^2 = 1) to both sides of the equation:
(x - 3)^2 + (y^2 + 2y + 1) = 24 + 1
(x - 3)^2 + (y + 1)^2 = 25
So, the equation of the circle is (x - 3)^2 + (y + 1)^2 = 25.
2. Find the center and radius of the circle by comparing the equation to the standard form equation: (x - h)^2 + (y - k)^2 = r^2.
Comparing the equations, we see that the center of the circle is (h, k) = (3, -1) and the radius (r) is sqrt(25) = 5.
3. Determine the equation of the line passing through the center of the circle (3, -1) and the given point (8, -2).
To find the equation of the line passing through these two points, we need to find the slope (m) and the y-intercept (b). The slope is given by the formula m = (y2 - y1) / (x2 - x1).
Using the coordinates (3, -1) and (8, -2), we have:
m = (-2 - (-1)) / (8 - 3)
m = (-2 + 1) / (8 - 3)
m = -1 / 5
Using the slope-intercept form of a linear equation, y = mx + b, and substituting the values of m and one of the points (3, -1), we can find the value of b:
-1 = (-1/5)(3) + b
-1 = -3/5 + b
b = -1 + 3/5
b = -2/5
So, the equation of the line passing through the center of the circle and the point (8, -2) is y = (-1/5)x - 2/5.
4. Find the equation of the diameter by finding the line perpendicular to the line we found in step 3 and passing through the center of the circle.
The equation of a line perpendicular to y = mx + b has a slope that is the negative reciprocal of m. In this case, the slope of the line we found in step 3 is -1/5, so the slope of the perpendicular line is 5.
Using the point-slope form of a linear equation, y - y1 = m(x - x1), and substituting the slope (m = 5) and the coordinates of the center of the circle (3, -1), we can find the equation of the diameter:
y - (-1) = 5(x - 3)
y + 1 = 5x - 15
y = 5x - 16
Hence, the equation of the diameter of the given circle, which passes through the point (8, -2), is y = 5x - 16.