Question

which point lies on the circle represented by the equation ((x - 3)^2 + (y + 4)^2 = 6^2)?

a. ((9, -2))

b. ((0, 11))

c. ((3, 10))

d. ((-9, 4))

e. ((-3, -4))

Explanation:

Step1: Recall circle equation rule

A point $(x,y)$ lies on the circle $(x-h)^2+(y-k)^2=r^2$ if substituting $x,y$ satisfies the equation.

Step2: Test Option A $(9,-2)$

Substitute $x=9, y=-2$:
$$(9-3)^2+(-2+4)^2 = 6^2+2^2 = 36+4=40$$
$6^2=36$, $40≠36$ → Wait, correction: Re-calculate:
Wait, $(-2+4)=2$, $2^2=4$; $(9-3)=6$, $6^2=36$. $36+4=40≠36$. Re-test all options:

Step2: Test Option E $(-3,-4)$

Substitute $x=-3, y=-4$:
$$(-3-3)^2+(-4+4)^2 = (-6)^2+0^2=36+0=36=6^2$$
This satisfies the equation.

Step3: Verify other options

Option B $(0,11)$:
$$(0-3)^2+(11+4)^2=9+225=234≠36$$
Option C $(3,10)$:
$$(3-3)^2+(10+4)^2=0+196=196≠36$$
Option D $(-9,4)$:
$$(-9-3)^2+(4+4)^2=144+64=208≠36$$
Option A $(9,-2)$:
$$(9-3)^2+(-2+4)^2=36+4=40≠36$$
Only Option E satisfies the equation.

Corrected Answer:

E. (-3,-4)

Step1: Define circle condition

A point $(x,y)$ is on the circle if substituting into the equation makes both sides equal.

Step2: Substitute Option E $(-3,-4)$

$$(-3-3)^2+(-4+4)^2 = (-6)^2 + 0^2 = 36 + 0 = 36 = 6^2$$

Step3: Confirm equality

The left-hand side equals the right-hand side, so the point lies on the circle.

Answer:

A. (9,-2)