Tangent lines, circle equations, centers, and radii

[Guest]

i have two problems that i am having an immense about of difficulty on both of them include tangent lines in relation to a circle. the first question is:

For the circle x^2 + y^2 + 6x - 4y + 3 = 0, Find:
a) the center and radius
which i figured out is center (-3, 2) radius square root of 10
b) the equation of the tangent at (-2, 5) which i have no idea how to do

my other question is:
a circle is tangent to the y axis at y = 3 and has one x-intercept at x = 1.
a) determine the other x-intercept
b) deduce the equation of the circle

i don't know how to begin this one at all. thank you in advance for your assistance

 

[pka]

For b).
The slope of the line through (-3,2) and (-2,5) is 3, so the slope of the tangent at (-2,5) is –1/3.

 

[soroban]

Re: Tangent lines

Hello, anh040216!

Here's the second one . . .

A circle is tangent to the y-axis at $\displaystyle y\,=\,3$ and has one x-intercept at $\displaystyle x\,=\,1$

a) Determine the other x-intercept
b) Deduce the equation of the circle

First, make a sketch . . .

        |
        |       * * *
        |   *           *
        | *               *
        |*                 *
        |
        *         C         *
      A o - - - - o (x,3)   *
   (0,3)*        / \        *
        |       /   \ 
        |*     /     \     *
        | * B /       \   *
    - - + - o - - - - - o - - -
        | (1,0) * * *   D
        |

The circle is tangent at $\displaystyle A(0,3)$
The center is at $\displaystyle C(x,3)$
An x-intercept is: $\displaystyle B(1,0)$

We know that: $\displaystyle \,AC\,=\,x\,=\,r\;$(radius) [1]

We know that: $\displaystyle \,BC\,=\,\sqrt{(x-1)^2\,+\,3^2}\,=\,r\;$(radius) [2]

Equate [1] and [2]: $\displaystyle \,x\;=\;\sqrt{(x-1)^2\;+\;9}$

Square both sides: $\displaystyle \,x^2\;= \;x^2\,-\,2x\,+\,1\,+\,9$

$\displaystyle \;\;$ and we have: $\displaystyle \,2x \,= \,10\;\;\Rightarrow\;\;x\,=\,5$

We have a circle with center $\displaystyle C(5,3)$ and radius $\displaystyle 5.$

Therefore, (b) the equation is: $\displaystyle \,(x\,-\,5)^2\,+\,(y\,-\,3)^2\;=\;25$


The x-intercepts occur where $\displaystyle y\,=\,0.$

The equation becomes: $\displaystyle \,(x\,-\,5)^2\,+\,(0\,-\,3)^2\;=\;25$

$\displaystyle \;\;$and we have the quadratic: $\displaystyle \,x^2\,-\,10x\,+\,9\;=\;0$

$\displaystyle \;\;$which factors: $\displaystyle \,(x\,-\,1)(x\,-\,9)\;=\;0$

$\displaystyle \;\;$and has roots: $\displaystyle \,x\,=\,1,\;9$

Therefore, (a) the x-intercepts are: (1,0) and (9,0).