Respuesta :
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
[tex](x-h)^{2} +(y-k)^{2}=r^{2}[/tex]
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
[tex]d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}[/tex]
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
[tex](x+3)^{2} +(y-4)^{2}=r^{2}[/tex]
Find the distance (radius) between the center (-3,4) and (-6,2)
substitute in the formula of distance
[tex]r=\sqrt{(2-4)^{2}+(-6+3)^{2}}[/tex]
[tex]r=\sqrt{(-2)^{2}+(-3)^{2}}[/tex]
[tex]r=\sqrt{13}\ units[/tex]
The equation of the circle is equal to
[tex](x+3)^{2} +(y-4)^{2}=(\sqrt{13}){2}[/tex]
[tex](x+3)^{2} +(y-4)^{2}=13[/tex]
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
[tex](10+3)^{2} +(4-4)^{2}=13[/tex]
[tex](13)^{2} +(0)^{2}=13[/tex]
[tex]169=13[/tex] -----> 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
[tex](x-1)^{2} +(y-3)^{2}=r^{2}[/tex]
Find the distance (radius) between the center (1,3) and (2,6)
substitute in the formula of distance
[tex]r=\sqrt{(6-3)^{2}+(2-1)^{2}}[/tex]
[tex]r=\sqrt{(3)^{2}+(1)^{2}}[/tex]
[tex]r=\sqrt{10}\ units[/tex]
The equation of the circle is equal to
[tex](x-1)^{2} +(y-3)^{2}=(\sqrt{10}){2}[/tex]
[tex](x-1)^{2} +(y-3)^{2}=10[/tex]
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
[tex](11-1)^{2} +(5-3)^{2}=10[/tex]
[tex](10)^{2} +(2)^{2}=10[/tex]
[tex]104=10[/tex] -----> is not true
therefore
The point is not on the circle
The statement is false
