Let the circle S : 36x² + 36y² - 108x + 120y + C = 0 be such that it neither intersects nor touches

Q: Why do we use $r < |h|$ and $r < |k|$? Can't we just say $r < h$?

That is a fantastic question! Remember that the center $(h, k)$ can be in any quadrant. The distance to the y-axis is $|h|$ and to the x-axis is $|k|$. If we ignore the absolute value, we might miss cases where the center has negative coordinates. Always use the absolute value to ensure the distance is positive.

Q: Why is the condition for a point inside the circle $S(x_1, y_1) < 0$?

Think of the circle equation as a function $S(x, y) = x^2 + y^2 + 2gx + 2fy + c = 0$. On the boundary, $S = 0$. Outside, $S > 0$. Inside, $S < 0$. It is a fundamental property of the circle's power. If you are ever unsure, test a point like the center $(0, 0)$ if the circle is $x^2 + y^2 - r^2 = 0$; you will see it gives $-r^2$, which is negative!

Q: Why did we divide the whole equation by 36 at the start?

To get the standard form $x^2 + y^2 + 2gx + 2fy + c = 0$. Without this, the coefficients of $x^2$ and $y^2$ are not 1, and the standard formulas for center and radius will give you incorrect results. Always normalize the equation first!

Question

Let the circle $S : 36x^2 + 36y^2 - 108x + 120y + C = 0$ be such that it neither intersects nor touches the co-ordinate axes. If the point of intersection of the lines, $x - 2y = 4$ and $2x - y = 5$ lies inside the circle $S$, then :

Options:

(A) $\frac{25}{9} < C < \frac{13}{3}$

(B) $100 < C < 165$

(C) $81 < C < 156$

(D) $100 < C < 156$

Sigma Percentile · Accepted Answer

### Analyzing the Setup Welcome, fellow traveler of the mathematical landscape. Today, we are not just solving a problem; we are exploring the delicate balance of a circle in the Cartesian plane. We have been given the equation: $$36x^2 + 36y^2 - 108x + 120y + C = 0$$ At first glance, it looks like a cluttered room. It is messy, unorganized, and frankly, a bit intimidating. But in mathematics, as in life, the first step to clarity is organization. ### Phase 1: The Art of Standardization We cannot work with an equation where the coefficients of $x^2$ and $y^2$ are 36. We need them to be 1. This is the standard form, the language in which circles speak. By dividing the entire equation by 36, we transform the chaos into order: $$x^2 + y^2 - 3x + \frac{10}{3}y + \frac{C}{36} = 0$$ Suddenly, the fog lifts. We can now identify the center and the radius. The center $C_0$ is given by $(-g, -f)$, which leads us to $(\frac{3}{2}, -\frac{5}{3})$. The radius $r$ is the square root of $g^2 + f^2 - c$. After a bit of careful arithmetic, we find: $$r = \frac{\sqrt{181 - C}}{6}$$ This radius is our key; it holds the secret to the circle's size, which is tied directly to our unknown constant $C$. ### Phase 2: The No-Go Zone The problem gives us a beautiful geometric constraint: the circle neither intersects nor touches the coordinate axes. Imagine the circle floating in the fourth quadrant. If it were to touch the y-axis, its radius would have to be exactly equal to the horizontal distance from the center to the y-axis, which is $|h|$. Since it does not touch, the radius must be strictly smaller than this distance. The same logic applies to the x-axis and the vertical distance $|k|$. Thus, we arrive at two critical inequalities: $r < |h|$ and $r < |k|$. Applying these, we set: $$\frac{\sqrt{181 - C}}{6} < \frac{3}{2} \quad ext{and} \quad \frac{\sqrt{181 - C}}{6} < \frac{5}{3}$$ Solving these inequalities is a test of patience and precision. For the first, we find $C > 100$. For the second, we find $C > 81$. Since both must be true, we take the stricter condition: **$C > 100$**. We have successfully fenced in our circle! ### Phase 3: The Intersection Point Now, let us turn our attention to the two lines: $x - 2y = 4$ and $2x - y = 5$. These lines are like two paths crossing in the woods. We need to find the exact spot where they meet. Solving these simultaneously is a classic exercise in linear algebra. From the first, $x = 2y + 4$. Substituting this into the second: $$2(2y + 4) - y = 5 \Rightarrow 3y = -3 \Rightarrow y = -1$$ Plugging this back, we find $x = 2$. Our intersection point $P$ is $(2, -1)$. ### Phase 4: The Final Trap The problem states that this point $P(2, -1)$ lies inside the circle. This is the final piece of the puzzle. In coordinate geometry, for a point $(x_1, y_1)$ to lie inside a circle $S(x, y) = 0$, the value of the expression $S(x_1, y_1)$ must be strictly less than zero. We substitute $x=2$ and $y=-1$ into our original equation: $$36(2)^2 + 36(-1)^2 - 108(2) + 120(-1) + C < 0$$ Let us calculate this with care: $$144 + 36 - 216 - 120 + C < 0$$ $$180 - 336 + C < 0$$ $$C - 156 < 0 \Rightarrow C < 156$$ ### Conclusion: The Synthesis We have arrived at the finish line. We have two bounds for $C$: from the axes constraint, $C > 100$, and from the point-inside-circle constraint, $C < 156$. Combining these, we get the final range: **$100 < C < 156$** You have navigated the algebra, respected the geometry, and solved the mystery. This is the essence of JEE Advanced—not just calculating, but understanding the story the numbers are telling you. Well done!

Animated Solution for Mathematics - Circles: Let the circle $S : 36 x^{2} + 36 y^{2} - 108 x + 120 y + C = 0$ be such that it neither intersects nor touches the co-ordinate axes. If the point of intersection of the lines, $x - 2 y = 4$ and $2 x - y = 5$ lies inside the circle $S$ , then :

Select Answer:

Visualized Solution

Standardizing the Circle Equation

Finding Center and Radius

Constraint: No Intersection with Axes

Solving for $C$ (Part 1)

Solving for $C$ (Part 2)

Finding the Point of Intersection

Calculating the Intersection Point

Point Inside the Circle Constraint

Substituting Point into Circle Equation

Solving for $C$ (Part 3)

Final Range of $C$

The Sigma Insight: Position of a Point with Respect to a Circle

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Phase 4

Conclusion

Similar Questions

If a variable line, $3 x + 4 y - λ = 0$ is such that the two circles $x^{2} + y^{2} - 2 x - 2 y + 1 = 0$ and $x^{2} + y^{2} - 18 x - 2 y + 78 = 0$ are on its opposite sides, then the set of all values of $λ$ is the interval :-

The straight line $2 x - 3 y = 1$ divides the circular region $x^{2} + y^{2} \leq 6$ into two parts. If $S = {(2, 3/4), (5/2, 3/4), (1/4, - 1/4), (1/8, 1/4)}$ then the number of points (s) in $S$ lying inside the smaller part is

Let $r_{1}$ and $r_{2}$ be the radii of the largest and smallest circles, respectively, which pass through the point $(- 4, 1)$ and having their centres on the circumference of the circle $x^{2} + y^{2} + 2 x + 4 y - 4 = 0$ . If $\frac{r _{1}}{r _{2}} = a + b 2$ , then $a + b$ is equal to :

The set of values of $k$ for which the circle $C : 4 x^{2} + 4 y^{2} - 12 x + 8 y + k = 0$ lies inside the fourth quadrant and the point $(1, - \frac{1}{3})$ lies on or inside the circle $C$ is :

Comprehension Passage

Consider $M$ with $r = \frac{1025}{513}$ . Let $k$ be the number of all those circles $C_{n}$ that are inside $M$ . Let $l$ be the maximum possible number of circles among these $k$ circles such that no two circles intersect. Then

Consider $M$ with $r = \frac{( 2 ^{199} - 1 ) 2}{2 ^{198}}$ . The number of all those circles $D_{n}$ that are inside $M$ is

Choose the incorrect statement about the two circles whose equations are given below: $x^{2} + y^{2} - 10 x - 10 y + 41 = 0$ and $x^{2} + y^{2} - 16 x - 10 y + 80 = 0$

Let the maximum and minimum values of $(8 x - x^{2} - 12 - 4)^{2} + (x - 7)^{2}, x \in R$ be $M$ and $m$ , respectively. Then $M^{2} - m^{2}$ is equal to _________