Question

The radius of the circle 𝐶 is ___ .

Answer 1

Correct Answer: 1.5

Step 1: The equation of the circle and the parabola
We are given the equation of a circle: $$ (x - a)^2 + y^2 = r^2, $$
where $(a, 0)$ is the center of the circle and $r$ is the radius.
We are also given the equation of a parabola: $$ y^2 = 4 - x. $$
We aim to find the relationship between the circle and the parabola.
Step 2: Substituting the equation of the parabola into the circle equation
Substitute $y^2 = 4 - x$ into the circle's equation. We get: $$ (x - a)^2 + (4 - x) = r^2. $$
Expanding this, we get: $$ (x - a)^2 + 4 - x = r^2, $$
or equivalently: $$ x^2 - 2ax + a^2 + 4 - x = r^2. $$
Simplify further: $$ x^2 - x(2a + 1) + (a^2 + 4 - r^2) = 0. $$
This is a quadratic equation in $x$, which we will denote as equation (1): $$ x^2 - x(2a + 1) + (a^2 + 4 - r^2) = 0. $$
Step 3: Condition for a single solution
For the circle and the parabola to intersect at exactly one point, the discriminant of the quadratic equation must be zero. The discriminant $D$ for the quadratic equation $Ax^2 + Bx + C = 0$ is given by: $$ D = B^2 - 4AC. $$
In our case, the coefficients are: - $A = 1$, - $B = -(2a + 1)$, - $C = a^2 + 4 - r^2$.
Thus, the discriminant is: $$ D = (-(2a + 1))^2 - 4(1)(a^2 + 4 - r^2). $$
Simplifying: $$ D = (2a + 1)^2 - 4(a^2 + 4 - r^2). $$
Expanding both terms: $$ D = (4a^2 + 4a + 1) - 4(a^2 + 4 - r^2). $$
Simplify further: $$ D = 4a^2 + 4a + 1 - 4a^2 - 16 + 4r^2. $$
This simplifies to: $$ D = 4r^2 + 4a - 15. $$
For the circle and the parabola to intersect at exactly one point, we set $D = 0$: $$ 4r^2 + 4a - 15 = 0. $$
Step 4: Solve for $a$ and $r$
From the equation: $$ 4r^2 + 4a - 15 = 0, $$
we can solve for $a$ in terms of $r$: $$ 4a = 15 - 4r^2, $$
$$ a = \frac{15 - 4r^2}{4}. $$
Step 5: Condition for $a \geq r$
We are given that $a \geq r$. So, we substitute the expression for $a$: $$ \frac{15 - 4r^2}{4} \geq r. $$
Multiply through by 4 to eliminate the denominator: $$ 15 - 4r^2 \geq 4r. $$
Rearrange the inequality: $$ 15 \geq 4r + 4r^2. $$
This simplifies to: $$ 4r^2 + 4r - 15 \leq 0. $$
Step 6: Solve the quadratic inequality
To solve the inequality $4r^2 + 4r - 15 \leq 0$, we first solve the corresponding quadratic equation: $$ 4r^2 + 4r - 15 = 0. $$
Using the quadratic formula: $$ r = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}, $$
where $A = 4$, $B = 4$, and $C = -15$, we get: $$ r = \frac{-4 \pm \sqrt{4^2 - 4(4)(-15)}}{2(4)} = \frac{-4 \pm \sqrt{16 + 240}}{8} = \frac{-4 \pm \sqrt{256}}{8} = \frac{-4 \pm 16}{8}. $$
Thus, the two solutions for $r$ are: $$ r = \frac{-4 + 16}{8} = \frac{12}{8} = \frac{3}{2}, $$ or $$ r = \frac{-4 - 16}{8} = \frac{-20}{8} = -\frac{5}{2}. $$
Since $r$ must be positive, we take $r = \frac{3}{2}$. 
Step 7: Conclusion
Therefore, the maximum possible value of $r$ is $\frac{3}{2}$, and since $a \geq r$, the radius of the circle $C$ is $\boxed{1.5}$.
 

Answer 2

Step 1: Equation of the Circle and Parabola
We are given the equation of the circle: $$ (x - a)^2 + y^2 = r^2, $$
and the equation of the parabola: $$ y^2 = 4 - x. $$
Step 2: Substitute the parabola equation into the circle equation
To solve the circle and parabola system, we substitute $y^2$ from the parabola equation into the circle equation. From the parabola, we know that $y^2 = 4 - x$, so we substitute this into the circle equation: $$ (x - a)^2 + (4 - x) = r^2. $$
Simplifying this equation: $$ (x - a)^2 + 4 - x = r^2. $$
Step 3: Expand and rearrange the equation
Now, expand $(x - a)^2$: $$ (x - a)^2 = x^2 - 2ax + a^2. $$
Substitute this back into the equation: $$ x^2 - 2ax + a^2 + 4 - x = r^2. $$
Simplify the equation: $$ x^2 - x(2a + 1) + (a^2 + 4 - r^2) = 0. $$
This is a quadratic equation in $x$. Let’s call this equation (1): $$ x^2 - x(2a + 1) + (a^2 + 4 - r^2) = 0. $$
Step 4: Apply the condition for tangency
For the circle and parabola to intersect at exactly one point, the discriminant ($D$) of the quadratic equation must be 0. The discriminant of a quadratic equation $Ax^2 + Bx + C = 0$ is given by $D = B^2 - 4AC$. In our case, $A = 1$, $B = -(2a + 1)$, and $C = a^2 + 4 - r^2$. So, we calculate the discriminant as: $$ D = (2a + 1)^2 - 4 \times 1 \times (a^2 + 4 - r^2). $$
Simplifying: $$ D = (2a + 1)^2 - 4(a^2 + 4 - r^2). $$
Expanding both terms: $$ D = (4a^2 + 4a + 1) - 4a^2 - 16 + 4r^2. $$
Simplify further: $$ D = 4a + 4r^2 - 15. $$
For tangency, we set $D = 0$: $$ 4a + 4r^2 - 15 = 0. $$
Thus: $$ 4r^2 + 4a - 15 = 0. $$
This is the condition for tangency between the circle and the parabola. Step 5: Solve for the maximum radius
We are given that $a \geq r$, so we substitute $a = r$ into the equation to find the maximum radius $r_{\text{max}}$: $$ 4r^2 + 4r - 15 = 0. $$
This is a quadratic equation in $r$. We solve it using the quadratic formula: $$ r = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}, $$
where $A = 4$, $B = 4$, and $C = -15$. Substituting these values, we get: $$ r = \frac{-4 \pm \sqrt{4^2 - 4 \times 4 \times (-15)}}{2 \times 4}. $$
Simplifying: $$ r = \frac{-4 \pm \sqrt{16 + 240}}{8}, $$
$$ r = \frac{-4 \pm \sqrt{256}}{8}, $$
$$ r = \frac{-4 \pm 16}{8}. $$
Thus, we have two possible solutions: $$ r = \frac{-4 + 16}{8} = \frac{12}{8} = \frac{3}{2}, $$
or $$ r = \frac{-4 - 16}{8} = \frac{-20}{8} = -\frac{5}{2}. $$
Since radius cannot be negative, we have: $$ r_{\text{max}} = \frac{3}{2}. $$
Step 6: Conclusion
Thus, the maximum radius of the circle is $\boxed{\frac{3}{2}}$.
The value of the radius of the circle $C$ is $\boxed{\frac{3}{2}}$.
From equation (1), we have $x^2 - 4x + 4 = 0$. Solving this quadratic equation: $$ x = 2. $$
Thus, $\alpha = 2$.