JEE Main 2023 (30 January Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Circles: Let $y = x + 2, 4 y = 3 x + 6$ and $3 y = 4 x + 1$ be three tangent lines to the circle $(x - h)^{2} + (y - k)^{2} = r^{2}$ . Then $h + k$ is equal to :

Q: Wait, why do we equate the distances to $L_2$ and $L_3$? What about $L_1$?

That is a great observation! The center $(h, k)$ is indeed equidistant from all three lines. However, to find the relationship between $h$ and $k$, we only need to equate the distances to any two lines. Equating the distances to $L_2$ and $L_3$ gives us a direct linear equation. If we needed the actual value of $r$, we would then use that relationship to find the distance to $L_1$.

Q: How do I handle the absolute value signs when equating the distances?

When you have an equation of the form $|A| = |B|$, it implies two possibilities: $A = B$ or $A = -B$. These two cases represent the two angle bisectors of the lines. In this problem, one case will lead to the center of the inscribed circle, and the other might lead to an excenter. Since we are looking for a specific value for $h+k$, we test the cases to see which one yields a valid result.

Select Answer:

Visualized Solution

Visualizing the Tangent Lines

Given tangent lines:
$L_{1} : x - y + 2 = 0$
$L_{2} : 3 x - 4 y + 6 = 0$
$L_{3} : 4 x - 3 y + 1 = 0$
Circle equation: $(x - h)^{2} + (y - k)^{2} = r^{2}$

The Equidistance Property

The center $(h, k)$ is equidistant from all three tangent lines.
This constant distance is exactly the radius $r$ of the circle.
Therefore, $d_{1} = d_{2} = d_{3} = r$ .

Distance Formula

The perpendicular distance $d$ from a point $(x_{1}, y_{1})$ to a line $a x + b y + c = 0$ is given by:
$d = \frac{∣ a x _{1} + b y _{1} + c ∣}{a ^{2} + b ^{2}}$

Distance from Center to $L_{2}$

Let's find the distance $d_{2}$ from $(h, k)$ to $L_{2} : 3 x - 4 y + 6 = 0$ .
$d_{2} = \frac{∣3 h - 4 k + 6∣}{3 ^{2} + ( - 4 ) ^{2}}$
$d_{2} = \frac{∣3 h - 4 k + 6∣}{5}$

Distance from Center to $L_{3}$

Now, find the distance $d_{3}$ from $(h, k)$ to $L_{3} : 4 x - 3 y + 1 = 0$ .
$d_{3} = \frac{∣4 h - 3 k + 1∣}{4 ^{2} + ( - 3 ) ^{2}}$
$d_{3} = \frac{∣4 h - 3 k + 1∣}{5}$

Equating Distances $d_{2}$ and $d_{3}$

Since both distances represent the radius $r$ , we can equate them:
$d_{2} = d_{3}$
$\frac{∣3 h - 4 k + 6∣}{5} = \frac{∣4 h - 3 k + 1∣}{5}$

Simplifying the Equation

We can cancel the common denominator $5$ from both sides:
$∣3 h - 4 k + 6∣ = ∣4 h - 3 k + 1∣$

Solving the Absolute Value (Case 1)

To solve the absolute value equation, we first consider the case where both expressions have the same sign:
$3 h - 4 k + 6 = 4 h - 3 k + 1$

Rearranging the Terms

Let's group the variables $h$ and $k$ on one side, and the constants on the other:
$6 - 1 = 4 h - 3 h + 4 k - 3 k$
$5 = h + k$

Final Conclusion

We have successfully found that $h + k = 5$ .
This perfectly matches Option 1.
Note: The negative case yields $h - k = - 1$ , which corresponds to excenters not listed in the simple integer options.

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The Sigma Insight: Equation of Tangent and Normal

Solution Diagram

Analyzing the Setup

Imagine you are standing on a vast, flat coordinate plane. Before you, three lines stretch out, intersecting to form a triangle. You are tasked with finding the heart of this triangle—the center of a circle that kisses all three lines perfectly.

The lines are defined as:

L_{1} : x - y + 2 = 0

L_{2} : 3 x - 4 y + 6 = 0

L_{3} : 4 x - 3 y + 1 = 0

Our goal is to find the coordinates of the center

(h, k)

of the circle

(x - h)^{2} + (y - k)^{2} = r^{2}

The Equidistance Property

The secret to this problem lies in the definition of a tangent. A line is tangent to a circle if and only if the perpendicular distance from the center of the circle to that line is exactly equal to the radius

r

This means our center

(h, k)

is a special point: it is equidistant from all three lines. Mathematically, this is expressed as

d_{1} = d_{2} = d_{3} = r

The Bridge

The Distance Formula

To calculate these distances, we reach for the perpendicular distance formula. For any line

a x + b y + c = 0

and a point

(x_{1}, y_{1})

, the distance

d

is given by:

d = \frac{∣ a x _{1} + b y _{1} + c ∣}{a ^{2} + b ^{2}}

Let us apply this to

L_{2}

and

L_{3}

. For

L_{2}

, the distance

d_{2}

is:

d_{2} = \frac{∣3 h - 4 k + 6∣}{3 ^{2} + ( - 4 ) ^{2}} = \frac{∣3 h - 4 k + 6∣}{5}

Similarly, for

L_{3}

, the distance

d_{3}

is:

d_{3} = \frac{∣4 h - 3 k + 1∣}{4 ^{2} + ( - 3 ) ^{2}} = \frac{∣4 h - 3 k + 1∣}{5}

The Absolute Value Dance

Now, we equate these distances because they both equal the radius

r

. We have:

\frac{∣3 h - 4 k + 6∣}{5} = \frac{∣4 h - 3 k + 1∣}{5}

The denominators cancel out, leaving us with the elegant equation:

∣3 h - 4 k + 6∣ = ∣4 h - 3 k + 1∣

Final Calculation

This is the moment of truth. We consider the case where the expressions inside the absolute value bars have the same sign:

3 h - 4 k + 6 = 4 h - 3 k + 1

Rearranging this, we get:

6 - 1 = 4 h - 3 h + 4 k - 3 k

This simplifies to the final relationship:

$h + k = 5$

Common Traps & FAQs

Wait, why do we equate the distances to

L_{2}

and

L_{3}

? What about

L_{1}

That is a great observation! The center

(h, k)

is indeed equidistant from all three lines. However, to find the relationship between

h

and

k

, we only need to equate the distances to any two lines. Equating the distances to

L_{2}

and

L_{3}

gives us a direct linear equation. If we needed the actual value of

r

, we would then use that relationship to find the distance to

L_{1}

How do I handle the absolute value signs when equating the distances?

When you have an equation of the form

∣ A ∣ = ∣ B ∣

, it implies two possibilities:

A = B

A = - B

. These two cases represent the two angle bisectors of the lines. In this problem, one case will lead to the center of the inscribed circle, and the other might lead to an excenter. Since we are looking for a specific value for

h + k

, we test the cases to see which one yields a valid result.

Why can't we just use energy conservation or some other physics principle here?

This is a pure coordinate geometry problem, not a physics problem! While the concept of 'equidistance' is similar to finding a potential equilibrium point, the tools here are strictly algebraic. We are using the properties of lines and circles in the Cartesian plane. Stick to the distance formula—it is your most powerful tool here.

Animated Solution for Mathematics - Circles: Let $y = x + 2, 4 y = 3 x + 6$ and $3 y = 4 x + 1$ be three tangent lines to the circle $(x - h)^{2} + (y - k)^{2} = r^{2}$ . Then $h + k$ is equal to :

Select Answer:

Visualized Solution

Visualizing the Tangent Lines

The Equidistance Property

Distance Formula

Distance from Center to $L_{2}$

Distance from Center to $L_{3}$

Equating Distances $d_{2}$ and $d_{3}$

Simplifying the Equation

Solving the Absolute Value (Case 1)

Rearranging the Terms

Final Conclusion

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Equidistance Property

The Bridge

The Absolute Value Dance

Final Calculation

Similar Questions

Let the lines $y + 2 x = 11 + 77$ and $2 y + x = 211 + 67$ be normal to a circle $C : (x - h)^{2} + (y - k)^{2} = r^{2}$ . If the line $11 y - 3 x = \frac{5 77}{3} + 11$ is tangent to the circle $C$ , then the value of $(5 h - 8 k)^{2} + 5 r^{2}$ is equal to ____.

The equations of the tangents drawn from the origin to the circle $x^{2} + y^{2} - 2 r x - 2 h y + h^{2} = 0$ , are

If the common tangent to the parabolas, $y^{2} = 4 x$ and $x^{2} = 4 y$ also touches the circle, $x^{2} + y^{2} = c^{2}$ , then $c$ is equal to:

If $y = m_{1} x + c_{1}$ and $y = m_{2} x + c_{2}, m_{1} \neq = m_{2}$ are two common tangents of circle $x^{2} + y^{2} = 2$ and parabola $y^{2} = x$ , then the value of $8∣ m_{1} m_{2} ∣$ is equal to

Let $T_{1}, T_{2}$ be two tangents drawn from $(- 2, 0)$ onto the circle $C : x^{2} + y^{2} = 1$ . Determine the circles touching $C$ and $T_{1}, T_{2}$ as their pair of tangents. Further, find the equations of all possible common tangents to these circles, when taken two at a time.

If a line, $y = m x + c$ is a tangent to the circle, $(x - 3)^{2} + y^{2} = 1$ and it is perpendicular to a line $L_{1}$ , where $L_{1}$ is the tangent to the circle, $x^{2} + y^{2} = 1$ at the point $(\frac{1}{2}, \frac{1}{2})$ ; then:

The tangent to the circle $C_{1} : x^{2} + y^{2} - 2 x - 1 = 0$ at the point (2, 1) cuts off a chord of length 4 from a circle $C_{2}$ whose centre is (3, -2). The radius of $C_{2}$ is :-

If the tangents drawn at the point $O (0, 0)$ and $P (1 + 5, 2)$ on the circle $x^{2} + y^{2} - 2 x - 4 y = 0$ intersect at the point $Q$ , then the area of the triangle $O P Q$ is equal to

Equation of a common tangent to the circle, $x^{2} + y^{2} - 6 x = 0$ and the parabola, $y^{2} = 4 x$ , is:

Animated Solution for Mathematics - Circles: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let y=x+2,4y=3x+6 and 3y=4x+1 be three tangent lines to the circle (x−h)2+(y−k)2=r2. Then h+k is equal to :

Select Answer:

Visualized Solution

Visualizing the Tangent Lines

The Equidistance Property

Distance Formula

Distance from Center to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L2​

Distance from Center to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L3​

Equating Distances .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } d2​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } d3​

Simplifying the Equation

Solving the Absolute Value (Case 1)

Rearranging the Terms

Final Conclusion

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Equidistance Property

The Bridge

The Absolute Value Dance

Final Calculation

Similar Questions

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the common tangent to the parabolas, y2=4x and x2=4y also touches the circle, x2+y2=c2, then c is equal to:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If y=m1​x+c1​ and y=m2​x+c2​,m1​=m2​ are two common tangents of circle x2+y2=2 and parabola y2=x, then the value of 8∣m1​m2​∣ is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The tangent to the circle C1​:x2+y2−2x−1=0 at the point (2, 1) cuts off a chord of length 4 from a circle C2​ whose centre is (3, -2). The radius of C2​ is :-

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the tangents drawn at the point O(0,0) and P(1+5​,2) on the circle x2+y2−2x−4y=0 intersect at the point Q, then the area of the triangle OPQ is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equation of a common tangent to the circle, x2+y2−6x=0 and the parabola, y2=4x, is:

Animated Solution for Mathematics - Circles: Let $y = x + 2, 4 y = 3 x + 6$ and $3 y = 4 x + 1$ be three tangent lines to the circle $(x - h)^{2} + (y - k)^{2} = r^{2}$ . Then $h + k$ is equal to :

Distance from Center to $L_{2}$

Distance from Center to $L_{3}$

Equating Distances $d_{2}$ and $d_{3}$

The equations of the tangents drawn from the origin to the circle $x^{2} + y^{2} - 2 r x - 2 h y + h^{2} = 0$ , are

If the common tangent to the parabolas, $y^{2} = 4 x$ and $x^{2} = 4 y$ also touches the circle, $x^{2} + y^{2} = c^{2}$ , then $c$ is equal to:

If $y = m_{1} x + c_{1}$ and $y = m_{2} x + c_{2}, m_{1} \neq = m_{2}$ are two common tangents of circle $x^{2} + y^{2} = 2$ and parabola $y^{2} = x$ , then the value of $8∣ m_{1} m_{2} ∣$ is equal to

The tangent to the circle $C_{1} : x^{2} + y^{2} - 2 x - 1 = 0$ at the point (2, 1) cuts off a chord of length 4 from a circle $C_{2}$ whose centre is (3, -2). The radius of $C_{2}$ is :-

If the tangents drawn at the point $O (0, 0)$ and $P (1 + 5, 2)$ on the circle $x^{2} + y^{2} - 2 x - 4 y = 0$ intersect at the point $Q$ , then the area of the triangle $O P Q$ is equal to

Equation of a common tangent to the circle, $x^{2} + y^{2} - 6 x = 0$ and the parabola, $y^{2} = 4 x$ , is: