Question

Question #f69b8

Answer 1

To show that a particular point is on the circle, simply substitute the coordinate values into the equation and see if the equation holds.

`x^2 + y^2 = 20`

`p = (2, -4)`

`=> 2^2 + (-4)^2 = 20`?

`=> 4 + 16 = 20`?

`=> 20 = 20`

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A line tangent to the circle at point `p` is perpendicular to the line connecting the point `p` and the center of the circle.

To find the equation of the tangent at point `p`, first we get the slope of the line connecting the point `p` and the center of the circle.

The center of the circle `x^2 + y^2 = 20` is a `(0, 0)`, So the slope `m'` is

`m' = (y_1 - y_2)/(x_1 - x_2)`

`=> m' = (0 - (-4))/(0 - 2)`

`=> m' = 4/-2`

`=> m' = -2`

To get the slope of the tangent line (which is perpendicular), we get the negative reciprocal of the slope we computed earlier.

`m = -1/(m')`

`=> m = -1/(-2)`

`=> m = 1/2`

Now let's get the y-intercept. The equation of a line in slope-intercept form is

`y = mx + b`

To get the y-intercept, let us substitute the coordinate values of an point that we know is on the line. For the tangent line, we know it passes through `(2, -4)`

`y = mx + b`

`=> -4 = (1/2)(2) + b`
`=> -4 = 1 + b`
`=> b = -3`

Therefore, the equation of the line tangent to the point (2, -4) is

`y = 1/2x - 3`

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If a line `k` is tangent to the circle at point `(k_x, k_y)` is parallel to another line `k'` that is tangent to the circle at point `(k_x', k_y')`, then either `k'` coincides with `k` (i.e. `k' = k`), or `k'` is on the exact other side of the circle. Let us leave the trivial coincidental lines case and focus on the other case.

What is the `(k_x, k_y)'s` relationship with `(k_x', k_y')`?
Since the parallel tangent lines are exactly on opposite sides of the circle, both points are endpoints of a circles diameter. We can use the midpoint formula to find `(k_x', k_y')`

`m_x = (k_x + k_x')/2`
`m_y = (k_y + k_y')/2`

Since the center of the circle is also the midpoint of the diameter, we have `m = (0, 0)`

`=> m_x = 0 = (2 + k_x')/2 = 2 + k_x'`

`=> k_x' = -2`

-

`=> m_y = 0 = (-4 + k_y')/2 = -4 + k_y'`

`=> k_y' = 4`

Hence, the other line is tangent to the circle at point `(-2, 4)`.
Since the other line is parallel to the first tangent line, the slope of the 2 lines should be the same.

`y = 1/2x + b`

Again, we need to find the `y`-intercept. Let's use the point of tangency to find it

`4 = 1/2(-2) + b`
`4 = -1 + b`
`b = 5`

Hence, the equation of the tangent line that is parallel to the tangent line at point `(2, -4)` is

`y = 1/2x + 5`