Answer:
Part 1) False
Part 2) False
Step-by-step explanation:
we know that
The equation of the circle in standard form is equal to
![(x-h)^{2} +(y-k)^{2}=r^{2}](https://tex.z-dn.net/?f=%28x-h%29%5E%7B2%7D%20%2B%28y-k%29%5E%7B2%7D%3Dr%5E%7B2%7D)
where
(h,k) is the center and r is the radius
In this problem the distance between the center and a point on the circle is equal to the radius
The formula to calculate the distance between two points is equal to
Part 1) given the center of the circle (-3,4) and a point on the circle (-6,2), (10,4) is on the circle.
true or false
substitute the center of the circle in the equation in standard form
![(x+3)^{2} +(y-4)^{2}=r^{2}](https://tex.z-dn.net/?f=%28x%2B3%29%5E%7B2%7D%20%2B%28y-4%29%5E%7B2%7D%3Dr%5E%7B2%7D)
Find the distance (radius) between the center (-3,4) and (-6,2)
substitute in the formula of distance
The equation of the circle is equal to
![(x+3)^{2} +(y-4)^{2}=(\sqrt{13}){2}](https://tex.z-dn.net/?f=%28x%2B3%29%5E%7B2%7D%20%2B%28y-4%29%5E%7B2%7D%3D%28%5Csqrt%7B13%7D%29%7B2%7D)
![(x+3)^{2} +(y-4)^{2}=13](https://tex.z-dn.net/?f=%28x%2B3%29%5E%7B2%7D%20%2B%28y-4%29%5E%7B2%7D%3D13)
Verify if the point (10,4) is on the circle
we know that
If a ordered pair is on the circle, then the ordered pair must satisfy the equation of the circle
For x=10,y=4
substitute
![(10+3)^{2} +(4-4)^{2}=13](https://tex.z-dn.net/?f=%2810%2B3%29%5E%7B2%7D%20%2B%284-4%29%5E%7B2%7D%3D13)
![(13)^{2} +(0)^{2}=13](https://tex.z-dn.net/?f=%2813%29%5E%7B2%7D%20%2B%280%29%5E%7B2%7D%3D13)
-----> is not true
therefore
The point is not on the circle
The statement is false
Part 2) given the center of the circle (1,3) and a point on the circle (2,6), (11,5) is on the circle.
true or false
substitute the center of the circle in the equation in standard form
![(x-1)^{2} +(y-3)^{2}=r^{2}](https://tex.z-dn.net/?f=%28x-1%29%5E%7B2%7D%20%2B%28y-3%29%5E%7B2%7D%3Dr%5E%7B2%7D)
Find the distance (radius) between the center (1,3) and (2,6)
substitute in the formula of distance
The equation of the circle is equal to
![(x-1)^{2} +(y-3)^{2}=(\sqrt{10}){2}](https://tex.z-dn.net/?f=%28x-1%29%5E%7B2%7D%20%2B%28y-3%29%5E%7B2%7D%3D%28%5Csqrt%7B10%7D%29%7B2%7D)
![(x-1)^{2} +(y-3)^{2}=10](https://tex.z-dn.net/?f=%28x-1%29%5E%7B2%7D%20%2B%28y-3%29%5E%7B2%7D%3D10)
Verify if the point (11,5) is on the circle
we know that
If a ordered pair is on the circle, then the ordered pair must satisfy the equation of the circle
For x=11,y=5
substitute
![(11-1)^{2} +(5-3)^{2}=10](https://tex.z-dn.net/?f=%2811-1%29%5E%7B2%7D%20%2B%285-3%29%5E%7B2%7D%3D10)
![(10)^{2} +(2)^{2}=10](https://tex.z-dn.net/?f=%2810%29%5E%7B2%7D%20%2B%282%29%5E%7B2%7D%3D10)
-----> is not true
therefore
The point is not on the circle
The statement is false