Question

If $4a^2 -5b^2 + 6a + 1 = 0$ and the line $ax + by + 1 = 0$ touches a fixed circle, then:

• A. centre of circle is at (3, 0)
• B. the radius of circle is $\sqrt{5}$
• C. the radius of circle is $\sqrt{3}$
• D. the circle passes through (1, 1)

Answer 1

Answer: (A), (B), (D)

(a), (b), (d)

Answer 2

The condition for the line $ax + by + 1 = 0$ to touch a circle with center $(h, k)$ and radius $r$ is given by the perpendicular distance from the center to the line being equal to the radius: $\frac{|ah + bk + 1|}{\sqrt{a^2 + b^2}} = r$ Squaring both sides, we get: $(ah + bk + 1)^2 = r^2 (a^2 + b^2)$ This equation must hold for all $a, b$ satisfying the constraint $4a^2 - 5b^2 + 6a + 1 = 0$.

Let's test option (a): center is at $(3, 0)$, so $h=3$ and $k=0$. Substituting these values into the touching condition: $(3a + 0b + 1)^2 = r^2 (a^2 + b^2)$ $(3a + 1)^2 = r^2 (a^2 + b^2)$ $9a^2 + 6a + 1 = r^2 a^2 + r^2 b^2$ From the constraint $4a^2 - 5b^2 + 6a + 1 = 0$, we can express $b^2$ in terms of $a$: $5b^2 = 4a^2 + 6a + 1 \implies b^2 = \frac{4a^2 + 6a + 1}{5}$ Substitute this expression for $b^2$ into the equation from the touching condition: $9a^2 + 6a + 1 = r^2 a^2 + r^2 \left(\frac{4a^2 + 6a + 1}{5}\right)$ Multiply by 5 to clear the fraction: $5(9a^2 + 6a + 1) = 5r^2 a^2 + r^2 (4a^2 + 6a + 1)$ $45a^2 + 30a + 5 = 5r^2 a^2 + 4r^2 a^2 + 6r^2 a + r^2$ $45a^2 + 30a + 5 = (9r^2) a^2 + (6r^2) a + r^2$ For this equation to hold for all valid values of $a$, the coefficients of the corresponding powers of $a$ on both sides must be equal. Comparing coefficients of $a^2$: $45 = 9r^2 \implies r^2 = 5$. Comparing coefficients of $a$: $30 = 6r^2 \implies r^2 = 5$. Comparing constant terms: $5 = r^2$. All comparisons consistently give $r^2 = 5$. This means that if the center is $(3, 0)$, the radius must be $\sqrt{5}$.

This confirms option (a) is consistent with option (b). So, the fixed circle has center $(3, 0)$ and radius $\sqrt{5}$.

Now let's check option (d): the circle passes through $(1, 1)$. The equation of the circle with center $(3, 0)$ and radius $\sqrt{5}$ is $(x-3)^2 + y^2 = (\sqrt{5})^2$, which simplifies to $(x-3)^2 + y^2 = 5$. Let's substitute the point $(1, 1)$ into this equation: $(1-3)^2 + (1)^2 = (-2)^2 + 1^2 = 4 + 1 = 5$ Since $5=5$, the point $(1, 1)$ lies on the circle. Thus, option (d) is also correct.

Option (c) states the radius is $\sqrt{3}$, which means $r^2=3$. This contradicts our finding that $r^2=5$. So, option (c) is incorrect.

Therefore, the correct statements are (a), (b), and (d).