Question

If the circle $x^2 + y^2 - 2x - 2y + 1 = 0$ is inscribed in a triangle whose two sides are co-ordinate axes and one side has negative slope cutting intercepts a and b on x and y axis, then :

• A. 1 = 1 - a + b
• B. 1 > a + b
• C. 1 < a + b
• D. $\sqrt{a^2 + b^2} = 1 - a + b$

Answer 1

Answer: (C)

1 < a + b

Answer 2

The equation of the circle is given by $x^2 + y^2 - 2x - 2y + 1 = 0$. By completing the square, we get $(x-1)^2 + (y-1)^2 = 1$. The center of the circle is $(1, 1)$ and the radius is $r=1$.

The triangle has sides along the coordinate axes ($x=0$, $y=0$) and a line with negative slope. The equation of this line is $\frac{x}{a} + \frac{y}{b} = 1$, or $bx + ay - ab = 0$. Since the circle is inscribed in the triangle and its center is in the first quadrant, the intercepts $a$ and $b$ must be positive. Also, for the triangle to contain the circle, $a$ and $b$ must be greater than the points of tangency on the axes, which are $(1,0)$ and $(0,1)$. Thus, $a > 1$ and $b > 1$.

The distance from the center of the inscribed circle to each side of the triangle must equal the radius. The distance from $(1,1)$ to $x=0$ is $|1|=1$, which equals $r$. The distance from $(1,1)$ to $y=0$ is $|1|=1$, which equals $r$. The distance from $(1,1)$ to $bx + ay - ab = 0$ must also be $1$: $\frac{|b(1) + a(1) - ab|}{\sqrt{b^2 + a^2}} = 1$ $|a + b - ab| = \sqrt{a^2 + b^2}$ Squaring both sides: $(a + b - ab)^2 = a^2 + b^2$ $a^2 + b^2 + a^2b^2 + 2ab - 2a^2b - 2ab^2 = a^2 + b^2$ $a^2b^2 + 2ab - 2a^2b - 2ab^2 = 0$ Since $a, b > 1$, we can divide by $ab$: $ab + 2 - 2a - 2b = 0$ Rearranging, we get: $ab - 2a - 2b + 4 = 2$ $(a-2)(b-2) = 2$ Since $a > 1$ and $b > 1$, for the product $(a-2)(b-2)$ to be positive, both factors must be positive or both must be negative. If $a-2 < 0$ and $b-2 < 0$, then $1 < a < 2$ and $1 < b < 2$. Let $a = 2-\epsilon$ where $0 < \epsilon < 1$. Then $(-\epsilon)(b-2) = 2 \implies b-2 = -2/\epsilon$, so $b = 2 - 2/\epsilon$. Since $\epsilon < 1$, $2/\epsilon > 2$, which implies $b < 0$. This contradicts $b > 1$. Therefore, we must have $a-2 > 0$ and $b-2 > 0$, which means $a > 2$ and $b > 2$.

Now let's examine the given options: (A) $1 = 1 - a + b \implies a = b$. This condition is met only when $a=b=2+\sqrt{2}$, but not for all valid $a,b$. (B) $1 > a+b$. Since $a>2$ and $b>2$, $a+b > 4$. Thus, $1 > a+b$ is false. (C) $1 < a+b$. Since $a>2$ and $b>2$, $a+b > 4$. Thus, $1 < a+b$ is always true. (D) $\sqrt{a^2+b^2} = 1 - a + b$. From $|a + b - ab| = \sqrt{a^2 + b^2}$, and since $a>2, b>2$, $a+b-ab = a+b-(2a+2b-2) = 2-a-b < 0$. So, $|a+b-ab| = -(a+b-ab) = ab-a-b$. Thus $\sqrt{a^2+b^2} = ab-a-b$. Equating this to the option: $ab-a-b = 1-a+b \implies ab-2b=1 \implies b(a-2)=1$. If $a=3/2$ (from earlier derivation attempt in option D), $b(3/2-2)=1 \implies b(-1/2)=1 \implies b=-2$, which contradicts $b>1$. Thus, this option is false.

The only statement that must be true is (C).