show that A(3,-1) lies on the circle x²+y²-2x+4y=0. Also find the otherend of of the diameter through A
Answers
Answer:
given that eq x²+y²-2x+4y=0
(3,-1). lies on it
i.e (3)^2+(-1)^2-2(3)+4(-1) =0
9+1-6-4=0
0=0
there fore, the point lies on the circle.
centre C (-g,-h)=(1,-2)
x+ 3/2=1. y-1/2=-2
hence ,x=-1
y=-3
(x,y)=(-1,-3
Given:
Equation of circle: x²+y²-2x+4y=0
Coordinates of A= (3, -1)
To find:
The other end of the diameter
Solution:
We can find the solution by following the given process-
We know that any point lying on the circumference will satisfy the equation of the circle.
So, A(3, -1) should satisfy x²+y²-2x+4y=0
Putting the values of A in the equation, we get
3²+(-1)²-2×3+4(-1)=0
9+1-6-4=0
10-10=0
So, point A lies on the circle.
Now, we know that the diameter of the circle passes through A, the center, and the other endpoint.
Let B be the other endpoint and its coordinates are (x, y).
The coordinates of the center can be determined from the equation of the circle.
To determine the coordinates, we will add 3 on both sides of the equation.
On adding 3, we get
x²+y²-2x+4y+1+2=3
(x²-2x+1)+(y²+4y+2)=3
(x-1)²+(y+√2)²=3
In the equation of circle (x-g)²+(y-h)²=r², the coordinates of the center are (g, h).
So, the coordinates of the center are (1, -√2).
The center of the circle is the mid-point of the diameter.
We will use the coordinates of A, center to find (x, y).
Using the midpoint formula, we get
(1, -√2)= (x+3)/2, (y-1)/2
(x+3)/2=1 and (y-1)/2= -√2
x+3=2 and y-1=-2√2
x= -1 and y=1-2√2
(x, y)= (-1, 1-2√2)
Therefore, the other end of the diameter is (-1, 1-2√2).