Question 765793
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Your description of the problem is confusing.  You actually have two separate problems since it is not possible for the two given lines to be tangent to the same circle centered at (-1,6) since the distance from one of your lines to the given center is *[tex \LARGE \sqrt{5}] and the distance from the other line to the given center is a very different *[tex \LARGE 3\sqrt{2}].


In addition, neither of your answers is correct.


Use the idea that the radius of a circle at the point where a line is tangent to the circle is perpendicular to the line.


Step 1:  Determine the slope of one of your given tangent lines.


Step 2:  Calculate the negative reciprocal of the slope from step 1.


Step 3:  Use the point-slope form of an equation of a line to determine an equation of the line that is perpendicular to the given line (the one you used in Step 1) and passes through the given center of the circle.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ -\ y_1\ =\ m(x\ -\ x_1) ]


where *[tex \Large \left(x_1,y_1\right)] are the coordinates of the given circle center and *[tex \Large m] is the slope calculated in step 2.


Step 4:  Use the given equation and the equation derived in step 3 as a 2X2 system of linear equations.  Solve the system to determine the point of intersection of the two lines.


Step 5:  Use the distance formula to calculate the distance from the center to the point of intersection of the two lines determined in step 4.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ d\ =\ \sqrt{(x_1\ -\ x_2)^2\ +\ (y_1\ -\ y_2)^2 }]


where *[tex \Large \left(x_1,y_1\right)] and *[tex \Large \left(x_2,y_2\right)] are the coordinates of the points for which you need the distance.


Step 6:  Using the coordinates of the center and the measure of the radius of the circle you just calculated in step 5, write the equation of the circle.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(x\,-\,h\right)^2\ +\ \left(y\,-\,k\right)^2\ =\ r^2]


where *[tex \Large \left(h,\,k\right)] are the coordinates of the given center of the circle and *[tex \Large r] is the radius calculated in step 5.


Repeat the process for the other line.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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