JEE Main 2016

LEVELJEE Advanced

Animated Solution for Mathematics - Circles: The centres of those circles which touch the circle, $x^{2} + y^{2} - 8 x - 8 y - 4 = 0$ , externally and also touch the x-axis, lie on:

Q: Why do we use $|k|$ instead of just $k$ for the radius?

A radius is a physical length, which must always be non-negative. If the circle lies below the x-axis, $k$ would be negative, so $|k|$ ensures the radius remains positive. Geometry doesn't care about signs, only magnitudes!

Q: Why does the distance between centers equal the sum of radii for external touching?

Imagine the two circles. If they touch externally, the line connecting their centers passes through the point of contact. The segment from the center of the first circle to the contact point is $r_1$, and from the contact point to the center of the second circle is $r$. Thus, the total distance is $r_1 + r$.

Q: How do I know it's a parabola just by looking at the final equation?

Look at the final equation. You have a squared term in $x$ (from $(x-4)^2$) and a linear term in $y$. Whenever you have one variable squared and the other linear, you are looking at the signature of a parabola. It is the geometric definition of a parabola in action!

Select Answer:

Visualized Solution

Equation of the Fixed Circle $C_{1}$

Given circle equation: $x^{2} + y^{2} - 8 x - 8 y - 4 = 0$
We need to find its center and radius.

Center and Radius of $C_{1}$

Comparing with $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$
Center $C_{1} = (- g, - f) = (4, 4)$
Radius $r_{1} = g^{2} + f^{2} - c = 16 + 16 - (- 4) = 6$

The Variable Circle $P$

Let the center of the required variable circle be $P (h, k)$ .
This circle touches the x-axis.

Radius of the Variable Circle

Since it touches the x-axis, the y-coordinate of the center determines the radius.
Radius $r = ∣ k ∣$

External Touching Condition

The variable circle touches the fixed circle $C_{1}$ externally.
Condition: Distance between centers $C_{1} P = r_{1} + r$

Setting up the Distance Equation

Distance $C_{1} P = (h - 4)^{2} + (k - 4)^{2}$
Sum of radii = $6 + ∣ k ∣$
Therefore, $(h - 4)^{2} + (k - 4)^{2} = 6 + ∣ k ∣$

Squaring Both Sides

To remove the square root, we square both sides.
$(h - 4)^{2} + (k - 4)^{2} = (6 + ∣ k ∣)^{2}$

Expanding the Terms

Expand the y-terms and the right side:
$(h - 4)^{2} + k^{2} - 8 k + 16 = 36 + 12∣ k ∣ + k^{2}$

Simplifying the Equation

Notice that $k^{2}$ appears on both sides.
Canceling $k^{2}$ :
$(h - 4)^{2} - 8 k + 16 = 36 + 12∣ k ∣$

Rearranging the Terms

Move all $k$ terms and constants to the right side:
$(h - 4)^{2} = 12∣ k ∣ + 8 k + 36 - 16$
$(h - 4)^{2} = 12∣ k ∣ + 8 k + 20$

Identifying the Locus

Replace $(h, k)$ with $(x, y)$ :
$(x - 4)^{2} = 12∣ y ∣ + 8 y + 20$
Case 1 ( $y \geq 0$ ): $(x - 4)^{2} = 20 (y + 1)$
Case 2 ( $y < 0$ ): $(x - 4)^{2} = - 4 (y - 5)$
Both cases represent a parabola.

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The Sigma Insight: Standard and General Equation of a Circle

Solution Diagram

Analyzing the Anchor

Every great journey begins with a solid foundation. We are given the equation of our fixed circle:

x^{2} + y^{2} - 8 x - 8 y - 4 = 0

To understand its properties, we complete the square to find its center and radius. By comparing this to the general form

x^{2} + y^{2} + 2 g x + 2 f y + c = 0

, we identify the center

C_{1} = (4, 4)

The radius

r_{1}

is calculated as follows:

r_{1} = g^{2} + f^{2} - c = 4^{2} + 4^{2} - (- 4) = 16 + 16 + 4 = 36 = 6

Our anchor is set: a circle at

(4, 4)

with a radius of

6

The Variable Dancer

Now, consider our variable circle with center

P (h, k)

. This circle is constrained to touch the x-axis.

Geometrically, the perpendicular distance from the center

(h, k)

to the x-axis must equal the radius. Therefore, the radius of our variable circle is

r = ∣ k ∣

The Geometric Kiss

The problem states that the variable circle touches the fixed circle externally. When two circles touch externally, the distance between their centers is exactly the sum of their radii.

Mathematically, this is expressed as:

C_{1} P = r_{1} + r

Substituting our known values into the distance formula, we obtain:

(h - 4)^{2} + (k - 4)^{2} = 6 + ∣ k ∣

This equation represents the soul of the problem, capturing the exact condition of contact.

The Algebraic Revelation

To solve for the locus, we square both sides to eliminate the radical:

(h - 4)^{2} + (k - 4)^{2} = (6 + ∣ k ∣)^{2}

Expanding both sides yields:

(h - 4)^{2} + k^{2} - 8 k + 16 = 36 + 12∣ k ∣ + k^{2}

Notice that the

k^{2}

terms on both sides cancel out. Simplifying the remaining expression, we get:

(h - 4)^{2} - 8 k + 16 = 36 + 12∣ k ∣

(h - 4)^{2} = 12∣ k ∣ + 8 k + 20

Replacing

(h, k)

with the general coordinates

(x, y)

, we arrive at the final locus equation:

(x - 4)^{2} = 12∣ y ∣ + 8 y + 20

Because the equation features a squared

x

term and a linear

y

term, the path of the center

P

is a parabola.

Common Traps & FAQs

Why do we use

∣ k ∣

instead of just

k

for the radius?

A radius is a physical length, which must always be non-negative. If the circle lies below the x-axis,

k

would be negative, so

∣ k ∣

ensures the radius remains positive. Geometry doesn't care about signs, only magnitudes!

Why does the distance between centers equal the sum of radii for external touching?

Imagine the two circles. If they touch externally, the line connecting their centers passes through the point of contact. The segment from the center of the first circle to the contact point is

r_{1}

, and from the contact point to the center of the second circle is

r

. Thus, the total distance is

r_{1} + r

How do I know it's a parabola just by looking at the final equation?

Look at the final equation. You have a squared term in

x

(from

(x - 4)^{2}

) and a linear term in

y

. Whenever you have one variable squared and the other linear, you are looking at the signature of a parabola. It is the geometric definition of a parabola in action!

Animated Solution for Mathematics - Circles: The centres of those circles which touch the circle, $x^{2} + y^{2} - 8 x - 8 y - 4 = 0$ , externally and also touch the x-axis, lie on:

Select Answer:

Visualized Solution

Equation of the Fixed Circle $C_{1}$

Center and Radius of $C_{1}$

The Variable Circle $P$

Radius of the Variable Circle

External Touching Condition

Setting up the Distance Equation

Squaring Both Sides

Expanding the Terms

Simplifying the Equation

Rearranging the Terms

Identifying the Locus

The Sigma Insight: Standard and General Equation of a Circle

Analyzing the Anchor

The Variable Dancer

The Geometric Kiss

The Algebraic Revelation

Similar Questions

The locus of the centre of a circle, which touches externally the circle $x^{2} + y^{2} - 6 x - 8 y + 14 = 0$ and also touches the y-axis, is given by the equation:

The locus of the centres of the circles, which touch the circle, $x^{2} + y^{2} = 1$ externally, also touch the y-axis and lie in the first quadrant, is :

A circle cuts a chord of length $4 a$ on the x-axis and passes through a point on the y-axis, distant $2 b$ from the origin. Then the locus of the centre of this circle, is:

A circle is given by $x^{2} + (y - 1)^{2} = 1$ , another circle $C$ touches it externally and also the x-axis, then the locus of its centre is

If the two circles $(x - 1)^{2} + (y - 3)^{2} = r^{2}$ and $x^{2} + y^{2} - 8 x + 2 y + 8 = 0$ intersect in two distinct point, then

If the two circles $(x - 1)^{2} + (y - 3)^{2} = r^{2}$ and $x^{2} + y^{2} - 8 x + 2 y + 8 = 0$ intersect in two distinct points, then

Let the centre of a circle, passing through the points $(0, 0), (1, 0)$ and touching the circle $x^{2} + y^{2} = 9$ , be $(h, k)$ . Then for all possible values of the coordinates of the centre $(h, k)$ , $4 (h^{2} + k^{2})$ is equal to

Let the circle $x^{2} + y^{2} = 4$ intersect $x$ -axis at the points $A (a, 0), a > 0$ and $B (b, 0)$ . Let $P (2 cos α, 2 sin α)$ , $0 < α < \frac{π}{2}$ and $Q (2 cos β, 2 sin β)$ be two points such that $(α - β) = \frac{π}{2}$ . Then the point of intersection of $A Q$ and $B P$ lies on :

If a circle C, whose radius is 3, touches externally the circle, $x^{2} + y^{2} - 4 y - 4 = 0$ at the point $(2, 2)$ , then the length of the intercept cut by this circle C, on the x-axis is equal to :

Circle(s) touching x-axis at a distance 3 from the origin and having an intercept of length $27$ on y-axis is (are)

Animated Solution for Mathematics - Circles: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The centres of those circles which touch the circle, x2+y2−8x−8y−4=0, externally and also touch the x-axis, lie on:

Select Answer:

Visualized Solution

Equation of the Fixed Circle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C1​

Center and Radius of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C1​

The Variable Circle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

Radius of the Variable Circle

External Touching Condition

Setting up the Distance Equation

Squaring Both Sides

Expanding the Terms

Simplifying the Equation

Rearranging the Terms

Identifying the Locus

The Sigma Insight: Standard and General Equation of a Circle

Analyzing the Anchor

The Variable Dancer

The Geometric Kiss

The Algebraic Revelation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The locus of the centre of a circle, which touches externally the circle x2+y2−6x−8y+14=0 and also touches the y-axis, is given by the equation:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The locus of the centres of the circles, which touch the circle, x2+y2=1 externally, also touch the y-axis and lie in the first quadrant, is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A circle cuts a chord of length 4a on the x-axis and passes through a point on the y-axis, distant 2b from the origin. Then the locus of the centre of this circle, 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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the two circles (x−1)2+(y−3)2=r2 and x2+y2−8x+2y+8=0 intersect in two distinct point, then

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the two circles (x−1)2+(y−3)2=r2 and x2+y2−8x+2y+8=0 intersect in two distinct points, then

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Circle(s) touching x-axis at a distance 3 from the origin and having an intercept of length 27​ on y-axis is (are)

Animated Solution for Mathematics - Circles: The centres of those circles which touch the circle, $x^{2} + y^{2} - 8 x - 8 y - 4 = 0$ , externally and also touch the x-axis, lie on:

Equation of the Fixed Circle $C_{1}$

Center and Radius of $C_{1}$

The Variable Circle $P$

The locus of the centre of a circle, which touches externally the circle $x^{2} + y^{2} - 6 x - 8 y + 14 = 0$ and also touches the y-axis, is given by the equation:

The locus of the centres of the circles, which touch the circle, $x^{2} + y^{2} = 1$ externally, also touch the y-axis and lie in the first quadrant, is :

A circle cuts a chord of length $4 a$ on the x-axis and passes through a point on the y-axis, distant $2 b$ from the origin. Then the locus of the centre of this circle, is:

A circle is given by $x^{2} + (y - 1)^{2} = 1$ , another circle $C$ touches it externally and also the x-axis, then the locus of its centre is

If the two circles $(x - 1)^{2} + (y - 3)^{2} = r^{2}$ and $x^{2} + y^{2} - 8 x + 2 y + 8 = 0$ intersect in two distinct point, then

If the two circles $(x - 1)^{2} + (y - 3)^{2} = r^{2}$ and $x^{2} + y^{2} - 8 x + 2 y + 8 = 0$ intersect in two distinct points, then

Circle(s) touching x-axis at a distance 3 from the origin and having an intercept of length $27$ on y-axis is (are)