Answer:
Possible point on y-axis are ( 0 , -5 ) and ( 0 , 11 )
Step-by-step explanation:
We are given coordinate of the center of the circle ( 6 , 3 ).
The Distance from the center of the circle to the edge = 10 units.
To find: Coordinate of the point on y-axis on the edge of the circle.
Given Distance is the Length of the radius of the circle.
Radius = 10 units
We know that standard form of the coordinate of the point on the y-axis.
Coordinate of the point on the y-axis = ( 0 , y )
Also, the distance formula of the two point is given as follows,
[tex]Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
Now, using distance formula
we have
[tex]\sqrt{(6-0)^2+(3-y)^2}=10[/tex]
[tex]\sqrt{36+3^2+y^2-6y}=10[/tex]
[tex]y^2-6y+45=100[/tex]
[tex]y^2-6y-55=0[/tex]
[tex]y^2-11y+5y-55=0[/tex]
y ( y - 11 ) + 5 ( y - 11 ) = 0
( y - 11 )( y + 5 ) = 0
y - 11 = 0 ⇒ y = 11
y + 5 = 0 ⇒ y = -5
Therefore, Possible point on y-axis are ( 0 , -5 ) and ( 0 , 11 )
Answer:
points (0, 11) and (0, -5)
Step-by-step explanation:
The equation of a circle is: (x - x0)^2 + (y - y0)^2 = r^2
where x0 is the x-coordinate of the center of the circle, y0 is the y-coordinate of the center of the circle and r is its radius. In this problem x0 = 6, y0 = 3 and r = 10 (the distance from the center of the circle to the edge). A plot of the circle is shown in the figure attached. There we can see that points (0, 11) and (0, -5) are, at the same time, on the y-axis and on the edge of the circle.
