A circle C of radius 1 is inscribed in...

MathematicsEquation of Tangent and NormalJEE Advanced 2008Difficult

Comprehension Passage

A circle

C

of radius 1 is inscribed in an equilateral triangle

P QR

. The points of contact of

C

with the sides

P Q, QR, R P

are

D, E, F

, respectively. The line

P Q

is given by the equation

3 x + y - 6 = 0

and the point

D

(\frac{3 3}{2}, \frac{3}{2})

. Further, it is given that the origin and the centre of

C

are on the same side of the line

P Q

Question 1:

The equation of circle $C$ is

(A)

(x - 23)^{2} + (y - 1)^{2} = 1

(B)

(x - 23)^{2} + (y + \frac{1}{2})^{2} = 1

(C)

(x - 3)^{2} + (y + 1)^{2} = 1

(D)

(x - 3)^{2} + (y - 1)^{2} = 1

Answer:D

Question 2:

Points $E$ and $F$ are given by

(A)

(\frac{3}{2}, \frac{3}{2}), (3, 0)

(B)

(\frac{3}{2}, \frac{1}{2}), (3, 0)

(C)

(\frac{3}{2}, \frac{3}{2}), (\frac{3}{2}, \frac{1}{2})

(D)

(\frac{3 3}{2}, \frac{3}{2}), (\frac{3}{2}, \frac{1}{2})

Answer:A

Question 3:

Equations of the sides $QR, R P$ are

(A)

y = \frac{2}{3} x + 1, y = - \frac{2}{3} x - 1

(B)

y = \frac{1}{3} x, y = 0

(C)

y = \frac{3}{2} x + 1, y = - \frac{3}{2} x - 1

(D)

y = 3 x, y = 0

Answer:D

Visualized Solution (Hindi)

Visualizing the Setup

Equilateral triangle $P QR$ with inscribed circle $C$ .
Radius of circle $C$ is $r = 1$ .
Equation of line $P Q$ : $3 x + y - 6 = 0$ .
Point of contact $D$ on $P Q$ : $(\frac{3 3}{2}, \frac{3}{2})$ .
Origin and center of $C$ are on the same side of $P Q$ .

Finding the Normal Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C D

The center $C$ lies on the normal to $P Q$ at point $D$ .
Slope of $P Q$ ( $m_{1}$ ) $= - 3$ .
Slope of normal $C D$ ( $m_{2}$ ) $= \frac{- 1}{m _{1}} = \frac{1}{3}$ .
Let the angle of the normal be $θ$ , then $tan θ = \frac{1}{3}$ .
This implies $cos θ = \frac{3}{2}$ and $sin θ = \frac{1}{2}$ .

Parametric Coordinates of Center .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C

Distance $C D = r = 1$ .
Using parametric form: $C = (x_{D} \pm r cos θ, y_{D} \pm r sin θ)$ .
$C = (\frac{3 3}{2} \pm 1 \cdot \frac{3}{2}, \frac{3}{2} \pm 1 \cdot \frac{1}{2})$ .
Case 1: $C_{1} = (23, 2)$ .
Case 2: $C_{2} = (3, 1)$ .

Applying the Origin Constraint

Line expression $L (x, y) = 3 x + y - 6$ .
For origin $(0, 0)$ : $L (0, 0) = - 6 < 0$ .
For $C_{1} (23, 2)$ : $L (23, 2) = 3 (23) + 2 - 6 = 6 + 2 - 6 = 2 > 0$ .
For $C_{2} (3, 1)$ : $L (3, 1) = 3 (3) + 1 - 6 = 3 + 1 - 6 = - 2 < 0$ .
Since $L (0, 0)$ and $L (C_{2})$ have the same sign, $C = (3, 1)$ .

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

Center $C = (3, 1)$ , Radius $r = 1$ .
Equation: $(x - 3)^{2} + (y - 1)^{2} = 1^{2}$ .
Final Equation: $(x - 3)^{2} + (y - 1)^{2} = 1$ .

Properties of the Equilateral Triangle

In an equilateral triangle, the incenter coincides with the centroid.
Contact points $D, E, F$ are the midpoints of sides $P Q, QR, R P$ respectively.
Distance from midpoint $D$ to vertices $P$ and $Q$ : $P D = D Q = r tan 6 0^{\circ} = 1 \cdot 3 = 3$ .

Finding Vertices .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q

Line $P Q$ has slope $- 3$ , so its angle is $12 0^{\circ}$ .
Direction cosines: $cos ϕ = - \frac{1}{2}$ , $sin ϕ = \frac{3}{2}$ .
Vertices $P, Q = (x_{D} \pm 3 cos ϕ, y_{D} \pm 3 sin ϕ)$ .
$P, Q = (\frac{3 3}{2} \pm \frac{3}{2}, \frac{3}{2} \mp \frac{3}{2})$ .
$P = (23, 0)$ and $Q = (3, 3)$ .

Finding Vertex .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R via Centroid

Centroid $C = (\frac{x _{P} + x _{Q} + x _{R}}{3}, \frac{y _{P} + y _{Q} + y _{R}}{3})$ .
$(3, 1) = (\frac{2 3 + 3 + x _{R}}{3}, \frac{0 + 3 + y _{R}}{3})$ .
$33 = 33 + x_{R} \Rightarrow x_{R} = 0$ .
$3 = 3 + y_{R} \Rightarrow y_{R} = 0$ .
Vertex $R = (0, 0)$ .

Coordinates of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } F

$E$ is the midpoint of $QR$ : $(\frac{3 + 0}{2}, \frac{3 + 0}{2}) = (\frac{3}{2}, \frac{3}{2})$ .
$F$ is the midpoint of $P R$ : $(\frac{2 3 + 0}{2}, \frac{0 + 0}{2}) = (3, 0)$ .
Points $E$ and $F$ are $(\frac{3}{2}, \frac{3}{2})$ and $(3, 0)$ .

Equations of Sides .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } QR and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R P

Equation of $QR$ (passes through $(0, 0)$ and $(3, 3)$ ): $y - 0 = \frac{3 - 0}{3 - 0} (x - 0) \Rightarrow y = 3 x$ .
Equation of $R P$ (passes through $(0, 0)$ and $(23, 0)$ ): $y = 0$ (The x-axis).

Summary and Key Takeaways

Circle Equation: $(x - 3)^{2} + (y - 1)^{2} = 1$
Points $E, F$ : $(\frac{3}{2}, \frac{3}{2}), (3, 0)$
Side Equations: $y = 3 x$ and $y = 0$
Key Property: In equilateral triangles, Incenter = Centroid and Contact Points = Midpoints.

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The circle $C_{1} : x^{2} + y^{2} = 3$ , with centre at $O$ , intersects the parabola $x^{2} = 2 y$ at the point $P$ in the first quadrant. Let the tangent to the circle $C_{1}$ at $P$ touches other two circles $C_{2}$ and $C_{3}$ at $R_{2}$ and $R_{3}$ , respectively. Suppose $C_{2}$ and $C_{3}$ have equal radii $23$ and centres $Q_{2}$ and $Q_{3}$ , respectively. If $Q_{2}$ and $Q_{3}$ lie on the $y$ -axis, then

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Comprehension Passage

A tangent

P T

is drawn to the circle

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at the point

P (3, 1)

. A straight line

L

, perpendicular to

P T

is a tangent to the circle

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The centre of the circle passing through $(0, 0)$ and $(1, 0)$ and touching the circle $x^{2} + y^{2} = 9$ is

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Comprehension Passage

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The equation of circle C is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Points E and F are given by

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equations of the sides QR,RP are

Visualized Solution (Hindi)

Visualizing the Setup

Finding the Normal Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } CD

Parametric Coordinates of Center .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C

Applying the Origin Constraint

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

Properties of the Equilateral Triangle

Finding Vertices .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q

Finding Vertex .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R via Centroid

Coordinates of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } F

Equations of Sides .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } QR and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } RP

Summary and Key Takeaways

Conceptually Similar Problems

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let P=(−1,0),Q=(0,0) and R=(3,33​) be three points. The equation of the bisector of the angle PQR is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let P=(−1,0),Q=(0,0) and R=(3,33​) be three points. Then the equation of the bisector of the angle PQR is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The equation of a circle with origin as a centre and passing through equilateral triangle whose median is of length 3a is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A circle is given by x2+(y−1)2=1, another circle C touches it externally and also the x-axis, then the locus of its centre is

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The centre of the circle passing through (0,0) and (1,0) and touching the circle x2+y2=9 is

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The equation of circle C is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Points E and F are given by

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equations of the sides QR,RP are

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let P=(−1,0),Q=(0,0) and R=(3,33​) be three points. The equation of the bisector of the angle PQR is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let P=(−1,0),Q=(0,0) and R=(3,33​) be three points. Then the equation of the bisector of the angle PQR is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The equation of a circle with origin as a centre and passing through equilateral triangle whose median is of length 3a is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A circle is given by x2+(y−1)2=1, another circle C touches it externally and also the x-axis, then the locus of its centre is

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Equations of the sides $QR, R P$ are

Finding the Normal Line $C D$

Parametric Coordinates of Center $C$

Equation of Circle $C$

Finding Vertices $P$ and $Q$

Finding Vertex $R$ via Centroid

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Equations of Sides $QR$ and $R P$

Let $P = (- 1, 0), Q = (0, 0)$ and $R = (3, 33)$ be three points. The equation of the bisector of the angle $P QR$ is

Let $P = (- 1, 0), Q = (0, 0)$ and $R = (3, 33)$ be three points. Then the equation of the bisector of the angle $P QR$ is

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A possible equation of $L$ is

The centre of the circle passing through $(0, 0)$ and $(1, 0)$ and touching the circle $x^{2} + y^{2} = 9$ is

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Let $P = (- 1, 0), Q = (0, 0)$ and $R = (3, 33)$ be three points. The equation of the bisector of the angle $P QR$ is

Let $P = (- 1, 0), Q = (0, 0)$ and $R = (3, 33)$ be three points. Then the equation of the bisector of the angle $P QR$ is

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