The general equation of a circle is
, where f, g and c are constants.
Put first point (0,0) in eqn, we got,
0 + 0 + 0 + 0 + c = 0
Now, the eqn becomes,
Put second point (4,0) in eqn, we got,
Now, the eqn becomes,
Put third point (3,3) in eqn, we got,
Now, the eqn becomes,
Complete the square for x and y terms,
so, centre = (2, 1) and radius
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
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 and are on the x-axis, we already know the perpendicular bisector of
. . It is the vertical line:
Now we seek a point on that vertical line
. . which is equidistant from and
. . We have: .
Therefore, the center is
You should be able to complete the problem now . . .