# Circle Question

• Aug 3rd 2007, 10:57 AM
charrie berri
Circle Question
Graph the circle (x+2)^2 + (y-1)^2= 25
• Aug 3rd 2007, 11:18 AM
earboth
Quote:

Originally Posted by charrie berri
Graph the circle (x+2)^2 + (y-1)^2= 25

Determine the equations of the tangents with slope of 3/4 =S

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:

$\displaystyle r: y=-\frac{4}{3}x-\frac{5}{3}$

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

$\displaystyle (x+2)^2+\left(-\frac{4}{3}x-\frac{5}{3}-1\right)^2 = 25$. After a few transformation you get:
$\displaystyle \frac{25}{9}x^2 + \frac{100}{9}x - \frac{125}{9}=0~\Longrightarrow~ x^2+4x-5=0$. Solve for x. Plug in the values into the equation of the straight line r. The tangent points are:
$\displaystyle T_1(1, -3),~~T_2(-5, 5)$

3. You know the slope of the tangent and one point of the tangent. Use the point-slope-formula.
$\displaystyle t_1:y = \frac{3}{4}x-\frac{15}{4}$

$\displaystyle t_2:y = \frac{3}{4}x+\frac{35}{4}$
• Aug 3rd 2007, 09:25 PM
earboth
Quote:

Originally Posted by charrie berri
Graph the circle (x+2)^2 + (y-1)^2= 25

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:

$\displaystyle t: y=\frac{3}{4}x + C$

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

$\displaystyle (x+2)^2+(\frac{3}{4}x + C-1)^2=25$ . Expand the brackets and after a few steps you'll get the equation:

$\displaystyle 25 \cdot x^2 + 8 \cdot x \cdot (3 \cdot C + 5) + 16C - 32C = 320$. Solve for x:

$\displaystyle x=\frac{4\left(\sqrt{-16C^2+80C+525}-3C-5\right)}{25}~~\vee~~$$\displaystyle ~~ x=-\frac{4\left(\sqrt{-16C^2+80C+525}-3C-5\right)}{25}$

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

$\displaystyle \sqrt{-16C^2+80C+525}=0$. Solve for C. You'll get:

$\displaystyle C=\frac{35}{4}~~\vee~~C=-\frac{15}{4}$ 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, 09:56 PM
earboth
Quote:

Originally Posted by charrie berri
Graph the circle (x+2)^2 + (y-1)^2= 25

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.