JEE Main 2019 (11 January)

LEVELJEE Main

Animated Solution for Mathematics - Circles: Two circles with equal radii are intersecting at the points $(0, 1)$ and $(0, - 1)$ . The tangent at the point $(0, 1)$ to one of the circles passes through the centre of the other circle. Then the distance between the centres of these circles is :

Q: Is the $T=0$ method always valid?

Absolutely! The $T=0$ method is a powerful tool in coordinate geometry for any conic section. It saves you the trouble of finding the slope and using the point-slope form, making your derivation much cleaner and less prone to calculation errors.

Select Answer:

Visualized Solution

Visualizing the Intersection Points

Intersection points: $A (0, 1)$ and $B (0, - 1)$
The common chord lies on the y-axis.

Locating the Centers by Symmetry

Circles have equal radii.
Centers lie on the perpendicular bisector of the common chord.
The perpendicular bisector is the x-axis.
Let centers be $C_{1} (h, 0)$ and $C_{2} (- h, 0)$ .

Calculating the Radius Squared $r^{2}$

Radius $r$ is the distance between $C_{1} (h, 0)$ and $A (0, 1)$ .
Using the distance formula: $r^{2} = (h - 0)^{2} + (0 - 1)^{2}$

Simplifying the Radius

$r^{2} = h^{2} + (- 1)^{2}$
$r^{2} = h^{2} + 1$

Equation of the First Circle

Standard form: $(x - x_{c})^{2} + (y - y_{c})^{2} = r^{2}$
Substitute center $(h, 0)$ and $r^{2} = h^{2} + 1$ :
$(x - h)^{2} + y^{2} = h^{2} + 1$

Applying the Tangent Formula $T = 0$

Equation of tangent at $(x_{1}, y_{1})$ is $T = 0$ .
For $(x - h)^{2} + y^{2} = r^{2}$ , tangent at $(0, 1)$ is:
$(0 - h) (x - h) + (1) (y) = h^{2} + 1$

Simplifying the Tangent Equation

Expand: $- h (x - h) + y = h^{2} + 1$
$- h x + h^{2} + y = h^{2} + 1$
Cancel $h^{2}$ : $- h x + y = 1$

Using the Second Center Condition

The tangent $- h x + y = 1$ passes through $C_{2} (- h, 0)$ .
Substitute $x = - h$ and $y = 0$ :
$- h (- h) + 0 = 1$

Solving for $h$

$- h (- h) = 1$
$h^{2} = 1$
Since $h$ represents a positive coordinate distance, $h = 1$ .

Final Distance Calculation

Distance between $C_{1} (h, 0)$ and $C_{2} (- h, 0)$ is $2 h$ .
Distance $= 2 (1) = 2$
Final Answer: 2

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The Sigma Insight: Equation of Tangent and Normal

Solution Diagram

Analyzing the Geometry

Imagine two identical circles symmetric about the

y

-axis. They intersect at the points

(0, 1)

and

(0, - 1)

Because the circles are mirror images across the

y

-axis, their centers must lie on the

x

-axis. Let the centers be located at

(h, 0)

and

(- h, 0)

Defining the Circle Equation

The radius squared,

r^{2}

, is the distance from the center

(h, 0)

to the intersection point

(0, 1)

. We calculate this as:

r^{2} = (h - 0)^{2} + (0 - 1)^{2} = h^{2} + 1

Consequently, the equation for the circle centered at

(h, 0)

is:

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

Applying the Tangent Condition

We utilize the

T = 0

method to find the equation of the tangent at the point

(0, 1)

. Substituting the point into the circle equation, we obtain:

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

Expanding and simplifying this expression yields:

- h x + h^{2} + y = h^{2} + 1

- h x + y = 1

Final Calculation

The problem implies that the tangent at

(0, 1)

passes through the center of the other circle, which is located at

(- h, 0)

. Substituting these coordinates into our tangent equation:

- h (- h) + 0 = 1

h^{2} = 1 \Rightarrow h = 1

Since the centers are at

(h, 0)

and

(- h, 0)

, the distance between them is

2 h

. Therefore, the distance between the centers is 2.

Common Traps & FAQs

Why are the centers at

(h, 0)

and

(- h, 0)

? Could they be anywhere else?

Great question! Because the circles have equal radii and share the same intersection points, the configuration must be symmetric about the perpendicular bisector of the common chord. Since the chord is on the y-axis, the perpendicular bisector is the x-axis. Any deviation would break the symmetry required for equal radii.

Is the

T = 0

method always valid?

Absolutely! The

T = 0

method is a powerful tool in coordinate geometry for any conic section. It saves you the trouble of finding the slope and using the point-slope form, making your derivation much cleaner and less prone to calculation errors.

Why did we get

h^{2} = 1

and not something else?

The condition that the tangent passes through the other center is a very specific geometric constraint. It forces a relationship between the radius and the distance of the centers. Solving

h^{2} = 1

simply reflects that the geometry is locked into a specific scale.

Animated Solution for Mathematics - Circles: Two circles with equal radii are intersecting at the points $(0, 1)$ and $(0, - 1)$ . The tangent at the point $(0, 1)$ to one of the circles passes through the centre of the other circle. Then the distance between the centres of these circles is :

Select Answer:

Visualized Solution

Visualizing the Intersection Points

Locating the Centers by Symmetry

Calculating the Radius Squared $r^{2}$

Simplifying the Radius

Equation of the First Circle

Applying the Tangent Formula $T = 0$

Simplifying the Tangent Equation

Using the Second Center Condition

Solving for $h$

Final Distance Calculation

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Geometry

Defining the Circle Equation

Applying the Tangent Condition

Final Calculation

Similar Questions

Two circles each of radius 5 units touch each other at the point $(1, 2)$ . If the equation of their common tangent is $4 x + 3 y = 10$ , and $C_{1} (α, β)$ and $C_{2} (γ, δ)$ , $C_{1} \neq = C_{2}$ are their centres, then $∣ (α + β) (γ + δ) ∣$ 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 :-

Two circles, each of radius 5 units, touch each other at $(1, 2)$ . If the equation of their common tangent is $4 x + 3 y = 10$ , find the equation of the circles.

The distance of the point $(6, - 22)$ from the common tangent $y = m x + c, m > 0$ , of the curves $x = 2 y^{2}$ and $x = 1 + y^{2}$ is

Circles with radii 3, 4 and 5 touch each other externally. If $P$ is the point of intersection of tangents to these circles at their points of contact, find the distance of $P$ from the points of contact.

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:

The straight line $x + 2 y = 1$ meets the coordinate axes at A and B. A circle is drawn through A, B and the origin. Then the sum of perpendicular distances from A and B on the tangent to the circle at the origin is :

Select Answer:

Visualized Solution

Visualizing the Intersection Points

Locating the Centers by Symmetry

Calculating the Radius Squared .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r2

Simplifying the Radius

Equation of the First Circle

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

Simplifying the Tangent Equation

Using the Second Center Condition

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } h

Final Distance Calculation

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Geometry

Defining the Circle Equation

Applying the Tangent Condition

Final Calculation

Similar Questions

.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; } Two circles, each of radius 5 units, touch each other at (1,2). If the equation of their common tangent is 4x+3y=10, find the equation of the circles.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The distance of the point (6,−22​) from the common tangent y=mx+c,m>0, of the curves x=2y2 and x=1+y2 is

.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:

Calculating the Radius Squared $r^{2}$

Applying the Tangent Formula $T = 0$

Solving for $h$

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 :-

Two circles, each of radius 5 units, touch each other at $(1, 2)$ . If the equation of their common tangent is $4 x + 3 y = 10$ , find the equation of the circles.

The distance of the point $(6, - 22)$ from the common tangent $y = m x + c, m > 0$ , of the curves $x = 2 y^{2}$ and $x = 1 + y^{2}$ is

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: