Let the tangents at the points A(4, -11) and B(8, -5) on the circle x² + y² - 3x + 10y - 15 = 0, int

Q: Wait, why do we use T=0 for the tangents? Can't we just find the slope of the radius and take the negative reciprocal?

You absolutely could! That is a valid geometric approach. However, the T=0 method is the 'JEE superpower.' It is a generalized algebraic shortcut that works for any conic section (circles, parabolas, ellipses, hyperbolas). Mastering T=0 saves you precious minutes in the exam hall because it bypasses the need to calculate slopes and derivatives. It is elegant, robust, and less prone to sign errors.

Q: Why is the line AB a tangent to the new circle? I thought AB was just a chord of the first circle.

That is a brilliant observation! You are correct—for the first circle, AB is indeed a chord. But the problem defines a new circle centered at C. In the context of this new circle, the problem explicitly states that AB is a tangent. We are not using the properties of the first circle for the second circle; we are using the geometric relationship defined by the problem statement. Always read the problem carefully to see which circle a property applies to!

Q: Is there a shortcut to finding the intersection point C?

In this specific case, the tangent at B turned out to be a vertical line (x=8), which made finding C very easy. If the tangents were not vertical, you would have a system of two linear equations. There is no 'magic' shortcut that avoids solving the system, but always look for vertical or horizontal lines first—they simplify the algebra significantly. If you see a coordinate like x=8, celebrate! It means the universe is giving you a break.

Question

Let the tangents at the points $A(4, -11)$ and $B(8, -5)$ on the circle $x^2 + y^2 - 3x + 10y - 15 = 0$, intersect at the point $C$. Then the radius of the circle, whose centre is $C$ and the line joining $A$ and $B$ is its tangent, is equal to

Options:

(A) $\frac{3\sqrt{3}}{4}$

(B) $2\sqrt{13}$

(C) $\sqrt{13}$

(D) $\frac{2\sqrt{13}}{3}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a vast, infinite coordinate plane. You have a circle, defined by the equation $x^2 + y^2 - 3x + 10y - 15 = 0$. It is a static, perfect shape.

We introduce two points, $A(4, -11)$ and $B(8, -5)$, resting on its boundary. These points are the anchors for our entire problem.

We are tasked with finding the radius of a new circle. This circle is centered at the intersection of the tangents drawn at $A$ and $B$, and it treats the line $AB$ as its tangent.

### The Power of T=0

When you see a tangent to a circle at a given point, the $T=0$ method is the most elegant tool in your arsenal. For any circle $x^2 + y^2 + 2gx + 2fy + c = 0$, the tangent at $(x_1, y_1)$ is given by:

$$xx_1 + yy_1 + g(x+x_1) + f(y+y_1) + c = 0$$

For point $A(4, -11)$, we substitute the values into our template. After simplifying, we arrive at the linear equation:

$$5x - 12y - 152 = 0$$

For point $B(8, -5)$, the calculation yields $13x - 104 = 0$, which simplifies to the vertical line:

$$x = 8$$

### The Intersection Point C

Now, we have two lines: $5x - 12y - 152 = 0$ and $x = 8$. Their intersection is the point $C$.

Substituting $x = 8$ into the first equation:

$$5(8) - 12y - 152 = 0$$

$$40 - 152 = 12y \Rightarrow 12y = -112$$

$$y = -\frac{28}{3}$$

Thus, our center $C$ is located at $(8, -\frac{28}{3})$.

### The Bridge of Line AB

We must define the line $AB$ to act as a tangent to this new circle. Using the two-point form, the slope $m$ is:

$$m = \frac{-5 - (-11)}{8 - 4} = \frac{6}{4} = \frac{3}{2}$$

Using the point-slope form, we derive the equation:

$$y - (-5) = \frac{3}{2}(x - 8)$$

$$2y + 10 = 3x - 24 \Rightarrow 3x - 2y - 34 = 0$$

### The Final Radius

The radius $r$ is the perpendicular distance from center $C(8, -\frac{28}{3})$ to the line $3x - 2y - 34 = 0$. We use the perpendicular distance formula:

$$r = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$$

Substituting our values:

$$r = \frac{|3(8) - 2(-\frac{28}{3}) - 34|}{\sqrt{3^2 + (-2)^2}}$$

The numerator simplifies as follows:

$$|24 + \frac{56}{3} - 34| = |\frac{56}{3} - 10| = |\frac{26}{3}|$$

The denominator is $\sqrt{9 + 4} = \sqrt{13}$. Thus:

$$r = \frac{26}{3\sqrt{13}}$$

Rationalizing the expression, we obtain the final result:

**$$r = \frac{2\sqrt{13}}{3}$$**

Select Answer:

Visualized Solution

Visualizing the Geometry

Tangent Equation $T = 0$

Tangent at $A (4, - 11)$

Simplifying Tangent at $A$

Tangent at $B (8, - 5)$

Intersection Point $C$

Equation of Line $A B$

The New Circle's Radius

Setting up the Distance Formula

Calculating the Final Radius

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Power of T=0

The Intersection Point C

The Bridge of Line AB

The Final Radius

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Select Answer:

Visualized Solution

Visualizing the Geometry

Tangent Equation .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T=0

Tangent at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A(4,−11)

Simplifying Tangent at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A

Tangent at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B(8,−5)

Intersection Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C

Equation of Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AB