Question

The equation of the circle which touches the axes of coordinates and the line $\frac{x}{3} + \frac{y}{4} = 1$ and whose centre lies in the first quadrant is $x^2 + y^2 + 2rx + 2ry + r^2 = 0$, then r can be equal to:

• A. 1
• B. 2
• C. 3
• D. 6

Answer 1

Answer: (A), (D)

1, 6

Answer 2

The given equation of the circle is $x^2 + y^2 + 2rx + 2ry + r^2 = 0$. Comparing this with the general equation of a circle $x^2 + y^2 + 2gx + 2fy + c = 0$, we have $g=r$, $f=r$, and $c=r^2$. The center of the circle is $(-g, -f) = (-r, -r)$. The radius of the circle is $\sqrt{g^2 + f^2 - c} = \sqrt{r^2 + r^2 - r^2} = \sqrt{r^2} = |r|$.

The problem states that the center of the circle lies in the first quadrant. This means both coordinates of the center must be positive: $-r > 0 \implies r < 0$. Since $r < 0$, the radius $|r|$ is equal to $-r$. Let $R$ be the radius, so $R = -r$. The center of the circle is then $(R, R)$ and the radius is $R$, where $R > 0$. This is the standard form of a circle touching both coordinate axes with its center in the first quadrant.

The circle also touches the line $\frac{x}{3} + \frac{y}{4} = 1$. We rewrite the equation of the line in the standard form $Ax + By + C = 0$. Multiplying by 12, we get $4x + 3y = 12$, which can be written as $4x + 3y - 12 = 0$.

The distance from the center of the circle $(R, R)$ to the line $4x + 3y - 12 = 0$ must be equal to the radius $R$. Using the distance formula $d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$: $R = \frac{|4R + 3R - 12|}{\sqrt{4^2 + 3^2}}$ $R = \frac{|7R - 12|}{\sqrt{16 + 9}}$ $R = \frac{|7R - 12|}{5}$

This implies $|7R - 12| = 5R$. Since $R$ is a radius, $R > 0$, so $5R$ must be positive. We consider two cases:

Case 1: $7R - 12 = 5R$ $2R = 12$ $R = 6$. This value of $R$ is positive, so it is a valid radius.

Case 2: $7R - 12 = -5R$ $12R = 12$ $R = 1$. This value of $R$ is also positive, so it is a valid radius.

We found two possible values for the radius $R$: $R=6$ and $R=1$. The problem states that the equation of the circle is $x^2 + y^2 + 2rx + 2ry + r^2 = 0$. We established that for the center to be in the first quadrant and the circle to touch the axes, the radius $R$ must be equal to $-r$. So, $r = -R$.

For $R=6$, we get $r = -6$. For $R=1$, we get $r = -1$.

The possible values for $r$ are $-6$ and $-1$. However, these values are not present in the given options: (A) 1, (B) 2, (C) 3, (D) 6.

This discrepancy suggests a potential typo in the question's equation. If the equation was intended to be $x^2 + y^2 - 2rx - 2ry + r^2 = 0$, then the center would be $(r, r)$ and the radius would be $|r|$. For the center to be in the first quadrant, $r > 0$. In this scenario, $r$ would directly represent the radius $R$. If we assume this intended form, then the possible values for $r$ (the radius) are $R=1$ and $R=6$. These values match the options provided.

Therefore, assuming the question implicitly refers to the radius $R$ and there's a sign error in the equation, the possible values for $r$ are 1 and 6.