1. The problem asks to explain why there are two possible values for $a$ in the equation $x^2 + y^2 = 100$ and to find these values. 2. The equation $x^2 + y^2 = 100$ represents a circle centered at the origin with radius $r = \sqrt{100} = 10$. 3. Given points $A(-4, 3)$ and $B(4, -3)$, we want to find values of $a$ such that the points $(a, y)$ satisfy the circle equation. 4. Substitute $x = a$ into the circle equation: $$a^2 + y^2 = 100$$ 5. Solve for $y^2$: $$y^2 = 100 - a^2$$ 6. For $y$ to be real, $100 - a^2 \geq 0$, so $a^2 \leq 100$ and $a$ lies between $-10$ and $10$. 7. Since $y = \pm \sqrt{100 - a^2}$, for each $a$ there are two possible $y$ values (positive and negative), explaining why there are two possible values for $a$ when considering points on the circle. 8. To find the specific values of $a$ for points $A$ and $B$, check their $x$-coordinates: - For $A(-4, 3)$, $a = -4$. - For $B(4, -3)$, $a = 4$. 9. Both $a = -4$ and $a = 4$ satisfy $a^2 = 16 \leq 100$, so both points lie on the circle. 10. To verify, substitute $a = \pm 4$ into the circle equation: $$(-4)^2 + 3^2 = 16 + 9 = 25 \neq 100$$ $$4^2 + (-3)^2 = 16 + 9 = 25 \neq 100$$ 11. Since $25 \neq 100$, points $A$ and $B$ do not lie on the circle $x^2 + y^2 = 100$. 12. However, the problem states two possible values for $a$; these are $a = \pm 8$ if we consider the point $(8, y)$ on the circle: $$8^2 + y^2 = 100 \Rightarrow 64 + y^2 = 100 \Rightarrow y^2 = 36 \Rightarrow y = \pm 6$$ 13. So the two points $(8, 6)$ and $(8, -6)$ lie on the circle, showing two possible $y$ values for $a=8$. 14. This explains why for a given $a$ there can be two possible $y$ values on the circle. 15. For part b), graphing the circle $x^2 + y^2 = 100$ and plotting points $(8, 6)$ and $(8, -6)$ confirms they lie on the circle. Final answer: The two possible values for $a$ are $a = 8$ and $a = -8$, each corresponding to two $y$ values $\pm 6$ on the circle $x^2 + y^2 = 100$.