Hey. Can somebody please help me out with this?
"A circle passes through the origin (0,0) and the points (4,0) and (3,3).
Find:
A. The centre of the circle
B. The equation to the tangent of the circle to the origin"
Cheers
The general equation of a circle is
$\displaystyle x^2+y^2+2fx+2gy+c=0$, where f, g and c are constants.
Put first point (0,0) in eqn, we got,
0 + 0 + 0 + 0 + c = 0
$\displaystyle \Rightarrow c = 0$
Now, the eqn becomes,
$\displaystyle x^2+y^2+2fx+2gy=0$
Put second point (4,0) in eqn, we got,
$\displaystyle (4)^2+(0)^2+2f(4)+2g(0)=0$
$\displaystyle \Rightarrow f = -2$
Now, the eqn becomes,
$\displaystyle x^2+y^2-4x+2gy=0$
Put third point (3,3) in eqn, we got,
$\displaystyle (3)^2+(3)^2-4(3)+2g(3)=0$
$\displaystyle \Rightarrow g = -1$
Now, the eqn becomes,
$\displaystyle x^2+y^2-4x-2y=0$
Complete the square for x and y terms,
$\displaystyle [x^2-2(2)x+2^2]-4+[y^2-2(1)y+1^2]-1=0$
$\displaystyle (x-2)^2+(y-1)^2-4-1=0$
$\displaystyle (x-2)^2+(y-1)^2=5$
so, centre = (2, 1) and radius $\displaystyle = \sqrt {5}$
B) Now, to find the slope of tangent, we can find the slope of the normal through centre (2, 1) to (0, 0)
slope of normal $\displaystyle = \frac{1-0}{2-0}=\frac{1}{2}$
so, slope of tangent = -2
now, find the equation of tangent line through (0, 0) having slope -2,
y = mx + b
0 = -2(0) + b
b = 0
so, eqn of tangent is
y = -2x
Please see attached graph
Hello, dungo!
Did you make a sketch?
A circle passes through the origin (0,0) and the points A(4,0) and B(3,3).
Find:
A. The centre of the circle
B. The equation to the tangent of the circle to the originCode:| : B | : o (3,3) | * * | * : * | * : * | * : * - * - - - + - - -o - - - |O : (4,0) | : A
Since $\displaystyle O$ and $\displaystyle B$ are on the x-axis, we already know the perpendicular bisector of $\displaystyle OA.$
. . It is the vertical line: $\displaystyle x = 2$
Now we seek a point on that vertical line $\displaystyle P(2,y)$
. . which is equidistant from $\displaystyle O$ and $\displaystyle B$
$\displaystyle OP^2 \:=\:2^2 + y^2\:=\:y^2 + 4$
$\displaystyle BP^2 \:=\:(2-3)^2 + (y-3)^2 \:=\: y^2 - 6y + 10$
. . We have: .$\displaystyle y^2+4 \:=\:y^2 - 6y + 10 \quad\Rightarrow\quad y \:=\:1$
Therefore, the center is $\displaystyle (2,1)$
You should be able to complete the problem now . . .