Question

Here is a circle, centre O, and the tangent to the circle at the point (8,15)

Answer 1

Answer:

y = -8/15x + 289/15

Explanation:

Given a circle with a radius of 17, and a centre at the origin (0,0), to find a line tangent to the point (8,15), there are a list of methods we could use to solve this problem. Instead of using derivatives (used to find the slope or gradient of a function at a given point), we can use the inverse tangent (goes from output to input, instead of input to output, but the inputs will still act as a function, which means you can only get one input for each output, this will get you from the side proportion to the angle, instead of angle to side proportion) as we know that a line connecting the origin to the point is always perpendicular to that line which is tangent to it. We can just take the opposite reciprocal (because when two lines are perpendicular, they always have opposite reciprocal slopes) of that, and then trace that slope back to the y-axis to find the it's y-intercept.

An opposite reciprocal just means to take that number, flip it's numerator, and denominator, and then make it negative.

So let's just use basic algebra to find the line going through points (0,0), and (8,15).

Because the line goes through the origin, we already know the y intercept is 0, so all we need to do is find the slope.

So slope [m] = (y2-y1)/(x2-x1).

Where (x1,y1) is the first point, and (x2, y2) is the second point. so m = 15-0/8-0 = 15/8.

Know that we have the slope of the first line, the perpendicular line will go through the same point, and have an opposite reciprocal -n^-1.

So 15/8 → 8/15 → -8/15 [this is the slope of the tangent].

Now since we know a point, and a slope, we can backtrack to find its y-intercept.

We can use point slope form:

y – y1 = m(x – x1)

(x1,y1) is the point we know [(8,15)], m is the slope we know [-8/15], and (x,y) is the point we are looking for. Also don't forget, since we are looking for a y-intercept, set x to zero because a y-intercept of anything is when x = 0.

Since we are only looking for y

y – y1 = m(x – x1) will become y = m(-x1) + y1.

This is because -y1 will cancel out when you add both sides by y1, and when x is 0, x1 will remain by itself in the form -x1.

So y = m(-x1) + y1 → y = (-8/15)(-8) + (15) →

64/15 + 15 = 64/15 + 15×15/15 = (64 + 225) / 15 = 289/15 [this is the y-intercept of the tangent line]

Now that we know the slope, and y- intercept of the tangent line, we can write the following equation in slope intercept:

m = -8/15, b = 289/15 → y = mx + b →

y = -8/15x + 289/15.

Answer 2

Answer:

y = -8/15x + 289/15