Graph the circle (x+2)^2 + (y-1)^2= 25

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- Aug 3rd 2007, 11:57 AMcharrie berriCircle Question
Graph the circle (x+2)^2 + (y-1)^2= 25

- Aug 3rd 2007, 12:18 PMearboth
Hello,

1. It is a circle with the centre C(-2, 1) and the radius 5.

2. The radius is perpendicular to the tangent in the tangent point. Thus the radius has the slope -4/3 and has to pass the centre:

Calculate the intersection of this line and the circle and you'll get the tangent points:

. After a few transformation you get:

. Solve for x. Plug in the values into the equation of the straight line r. The tangent points are:

3. You know the slope of the tangent and one point of the tangent. Use the point-slope-formula.

- Aug 3rd 2007, 10:25 PMearboth
Hello,

here is another approach:

You allways find 2 parallel tangents to a circle. The equation of these tangents differ only in the constant summand. With your values the tangents have the form:

Now calculate the intersection points of these tangents and the circle. Plug in the tangent term into the equation of the circle:

. Expand the brackets and after a few steps you'll get the equation:

. Solve for x:

You get only one intersetion point (that is the tangent point) if the radicand equals zero:

. Solve for C. You'll get:

Plug in this value into the equation of the parallel lines and you have the result you know from my previous post. - Aug 4th 2007, 10:56 PMearboth
Hello,

in my previous post I mentioned that the centre of the circle is C(-2, 1) and the radius is 5. I have some difficulties to imagine what difficulties you could have to draw a circle if all data are present :confused:

But nevertheles here it is:

By the way: Don't delete your problem when you have received an answer. Other members of the forum could benefit from your problem and the way it is solved.