A circle has the equation x^2+y^2-4x+6y-12=0
and i have already found that the centre has a coordinate of (2,-3), and a radius of 5 units. But i need to find points on the circle that has the gradient of -4/3 when a tangent is drawn on the point. How do i do that???
Assuming you have not done Calculus, we will work without it.
Let O = centre point of the circle
If the tangent to the point P has a gradient of m, what can you say about the gradient (slope) of the radius OP, denoted by p?
We know that OP and the tangent are perpendicular to each other, so m*p=-1.
You can therefore find p knowing m.
What is the line passing through the centre (2,-3) with a slope of p?
(y-(-3)) = p(x-2)
The intersection of this line with the circle will give you the required tangent points. Note that there are two such points. It will be the solution of a quadratic equation.
I see! Thanks!
You're welcome!
To find the points on the circle with a gradient of -4/3 when a tangent is drawn, we can use the following steps:
Step 1: Rewrite the equation of the circle in standard form by completing the square for both x and y.
Given equation: x^2 + y^2 - 4x + 6y - 12 = 0
Rearrange the equation:
(x^2 - 4x) + (y^2 + 6y) = 12
Complete the square for x:
(x^2 - 4x + 4) + (y^2 + 6y) = 12 + 4
(x - 2)^2 + (y^2 + 6y) = 16
Complete the square for y:
(x - 2)^2 + (y^2 + 6y + 9) = 16 + 9
(x - 2)^2 + (y + 3)^2 = 25
Step 2: Now we have the equation of the circle in standard form: (x - h)^2 + (y - k)^2 = r^2, where (h, k) is the center and r is the radius. In this case, the center of the circle is (2, -3), and the radius is 5 units.
Step 3: Find the gradient of a tangent line at any point on the circle.
The gradient, or slope, of a tangent line to a circle at a given point is the negative reciprocal of the gradient of the line connecting the center of the circle to that point.
The slope of the line connecting the center (2, -3) to any point (x, y) on the circle is given by:
slope = (y - k) / (x - h) = (y - (-3)) / (x - 2) = (y + 3) / (x - 2)
To find the points on the circle with a gradient of -4/3, we set the slope equal to -4/3:
(y + 3) / (x - 2) = -4/3
Step 4: Solve the equation for y in terms of x:
Cross-multiplying gives us:
3(y + 3) = -4(x - 2)
3y + 9 = -4x + 8
3y = -4x - 1
y = (-4/3)x - 1/3
So, the equation of the tangent line with a gradient of -4/3 is y = (-4/3)x - 1/3.
Step 5: Substitute this equation of the tangent line back into the equation of the circle to find the intersection points.
Substitute y = (-4/3)x - 1/3 into (x - 2)^2 + (y + 3)^2 = 25:
(x - 2)^2 + ((-4/3)x - 1/3 + 3)^2 = 25
Simplify and solve for x to get the x-coordinates of the points of intersection.
Expand:
(x - 2)^2 + ((-4/3)x + 8/3)^2 = 25
(x - 2)^2 + (16/9)x^2 + (64/9)x + 64/9 = 25
Combine like terms:
(x^2 - 4x + 4) + (16/9)x^2 + (64/9)x + 64/9 = 25
9x^2 - 36x + 36 + 16x^2 + 64x + 64 = 225
Combine like terms:
25x^2 + 28x + 100 = 225
Move 225 to the other side of the equation:
25x^2 + 28x + 100 - 225 = 0
25x^2 + 28x - 125 = 0
Now, you can solve this quadratic equation using the quadratic formula or factoring to find the values of x. Once you have the x-coordinates, you can substitute them back into the equation of the tangent line to find the corresponding y-coordinates.
Note: The quadratic equation may have zero, one, or two real solutions, depending on the discriminant. Each solution (x, y) represents a point on the circle with a tangent line with a gradient of -4/3.