## Active forum topics

- General Solution of $y' = x \, \ln x$
- engineering economics: construct the cash flow diagram
- Eliminate the Arbitrary Constants
- Law of cosines
- Maxima and minima (trapezoidal gutter)
- Special products and factoring
- Integration of 4x^2/csc^3x√sinxcosx dx
- application of minima and maxima
- Sight Distance of Vertical Parabolic Curve
- Application of Differential Equation: Newton's Law of Cooling

## New forum topics

- General Solution of $y' = x \, \ln x$
- engineering economics: construct the cash flow diagram
- Integration of 4x^2/csc^3x√sinxcosx dx
- Maxima and minima (trapezoidal gutter)
- Special products and factoring
- Newton's Law of Cooling
- Law of cosines
- Can you help me po to solve this?
- Eliminate the Arbitrary Constants
- Required diameter of solid shaft

## Recent comments

- 400000=120[14π(D2−10000)]

(…3 weeks 2 days ago - Use integration by parts for…1 month 3 weeks ago
- need answer1 month 3 weeks ago
- Yes you are absolutely right…1 month 3 weeks ago
- I think what is ask is the…1 month 3 weeks ago
- $\cos \theta = \dfrac{2}{…1 month 3 weeks ago
- Why did you use (1/SQ root 5…1 month 3 weeks ago
- How did you get the 300 000pi1 month 3 weeks ago
- It is not necessary to…1 month 3 weeks ago
- Draw a horizontal time line…1 month 4 weeks ago

## Create a circle of radius 1

Create a circle of radius 1 with center at (2, 1). The equation of the circle would be (x - 2)

^{2}+ (y - 1)^{2}= 1. Then draw a line through (7, -4) that is tangent to your circle, you can see that you can draw two such lines, one is above the (2, 1) and the other is below the point (2, 1). You only need to draw one of the lines. Call the point of tangency as (x_{1}, y_{1}) and our task now is to locate this point of tangency. You can find (x_{1}, y_{1}) knowing that this point is common to the circle and the line. Once you get (x_{1}, y_{1}), you have now two points on the line to be used to find its equation.## Here is the details of the

Here is the details of the solution:

$(x - 2)^2 + (y - 1)^2 = 1^2$

$(x^2 - 4x + 4) + (y^2 - 2y + 1) = 1$

$x^2 + y^2 - 4x - 2y + 4 = 0$

The circle above will pass through (x

_{1}, y_{1})${x_1}^2 + {y_1}^2 - 4x_1 - 2y_1 + 4 = 0$ ← Equation (1)

Equation of tangent to x

^{2}+ y^{2}- 4x - 2y + 4 = 0$xx_1 + yy_1 - 4\left( \dfrac{x + x_1}{2} \right) - 2\left( \dfrac{y + y_1}{2} \right) + 4 = 0$

$xx_1 + yy_1 - 2(x + x_1) - (y + y_1) + 4 = 0$

$7x_1 - 4y_1 - 2(7 + x_1) - (-4 + y_1) + 4 = 0$

$7x_1 - 4y_1 - 14 - 2x_1 + 4 - y_1 + 4 = 0$

$5x_1 - 5y_1 - 6 = 0$

$y_1 = x_1 - \frac{6}{5}$ ← Equation (2)

Substitute y

_{1}= x_{1}- 6/5 of Equation (2) to Equation (1)${x_1}^2 + (x_1 - \frac{6}{5})^2 - 4x_1 - 2(x_1 - \frac{6}{5}) + 4 = 0$

${x_1}^2 + ({x_1}^2 - \frac{12}{5}x_1 + \frac{36}{25}) - 4x_1 - 2x_1 + \frac{12}{5} + 4 = 0$

$2{x_1}^2 - \frac{42}{5}x_1 + \frac{196}{25} = 0$

$x_1 = \frac{14}{5} ~ \text{and} ~ \frac{7}{5}$

From Equation (2)

$y_1 = \frac{8}{5} ~ \text{and} ~ \frac{1}{5}$

$m = \dfrac{y_2 - y_1}{x_2 - x_1}$

$y - y_1 = m(x - x_1)$

For line L

_{1}$y - \frac{8}{5} = -\frac{4}{3}(x - \frac{14}{5})$

$15y - 24 = -20(x - \frac{14}{5})$

$15y - 24 = -20x + 56$

$20x + 15y - 80 = 0$

$4x + 3y - 16 = 0$

answerFor line L

_{2}$y - \frac{1}{5} = -\frac{3}{4}(x - \frac{7}{5})$

$20y - 4 = -15(x - \frac{7}{5})$

$20y - 4 = -15x + 21$

$15x + 20y - 25 = 0$

$3x + 4y - 5 = 0$

answer## I am happy to register this.

I am happy to register this.