Draw Concentric Circles One at a Time

In Geometry, the objects are said to be concentric when they share a common middle. Circles, spheres, regular polyhedra, regular polygons are concentric as they share the same eye point. In Euclidean Geometry , two concentric circles should take different radii from each other. In this article, y'all will learn what are concentric circles, the theorem on concentric circles, the region between the concentric circles, equations, and examples in detail.

Concentric Circles Pregnant

The circles with a common middle are known as concentric circles and have different radii. In other words, information technology is defined equally two or more circles that have the same centre point. The region between two concentric circles are of different radii is known equally an annulus.

Concentric Circle Equations

Let the equation of the circumvolve with middle (-g, -f) and radius √[g^two+f²-c] be

tenⁱⁱ + y² + 2gx + 2fy + c =0

Therefore, the equation of the circle concentric with the other circle be

ten² + y² + 2gx + 2fy + c' =0

It is observed that both the equations have the same centre (-g, -f), simply they accept different radii, where c≠ c'

Similarly, a circle with centre (h, k), and the radius is equal to r, then the equation becomes

( 10 – h )^two + ( y – k )ⁱⁱ = r²

Therefore, the equation of a circle concentric with the circumvolve is

( x – h )² + ( y – k )² = r_ane ^two

Where r ≠ r₁

By assigning different values to the radius in the to a higher place equation, nosotros shall get a family of circles.

Concentric Circles – Theorem

In two concentric circles, the chord of the larger circle, which touches the smaller circle, is bisected at the point of contact.

Proof

Given:

Consider two concentric circles C₁ and C₂, with center O and a chord AB of the larger circumvolve C₁, touching the smaller circumvolve C₂ at the betoken P equally shown in the effigy below.

Structure:

Join OP.

To bear witness: AP = BP

Proof:

Since AB is the chord of larger circle C₁, it becomes the tangent to C₂ at P.

OP is the radius of circle C_ii.

We know that the radius is perpendicular to the tangent at the signal of contact.

Then, OP ⊥ AB

Now AB is a chord of the circle C1 and OP ⊥ AB.

Therefore, OP is the bisector of the chord AB.

Thus, the perpendicular from the middle bisects the chord, i.e., AP = BP.

Region Between Concentric Circles

As mentioned above, the region betwixt two concentric circles is called the annulus. However, we tin can discover the perimeter and expanse of the annulus using appropriate formulas. The surface area of the annulus is calculated by subtracting the surface area of smaller circles from the area of the larger circle.

Suppose R is the radius of the larger circle and r is the radius of the smaller circle such that the area of the region divisional past these two circles is given past:

Expanse of annulus = πRⁱⁱ – πr^two

Learn more than well-nigh annulus here.

Concentric Circumvolve Examples

Question: Notice the equation of the circle concentric with the circle x² + yⁱⁱ + 4x – 8y – 6 =0, having the radius double of its radius.

Solution:

Given, circle equation: ten² + yⁱⁱ + 4x – 8y – 6 =0

We know that the equation of the circle is tenⁱⁱ + y² + 2gx + 2fy + c =0

From the given equation, the center point is (-ii, iv)

Therefore, the radius of the given equation volition be

r = √[g²+f²-c]

r = √[iv+16+6]

r = √26

Let R be the radius of the concentric circle.

It is given that, the radius of the concentric circle is double of its radius, then

R = 2r

R = 2√26

Therefore, the equation of the concentric circle with the radius R and the heart point (-g, -f ) is

( 10 – g )² + ( y – f )ⁱⁱ = Rⁱⁱ

(10 + 2)ⁱⁱ + ( y – 4 )ⁱⁱ = (2√26 )ⁱⁱ

x² + 4x + 4 + y² – 8y + xvi = 4 (26)

10² + y² + 4x – 8y +20 = 104

x² + y² + 4x – 8y – 84 = 0

Question 2:

Find the expanse between two concentric circles whose diameters are 35 cm and 21 cm.

Solution:

Given,

The diameters of the two circles are 35 cm and 21 cm.

So, R = 35/2 cm

r = 21/two cm

Expanse of betwixt concentric circles = πR² – πr²

= (22/seven) × (35/2) × (35/2) – (22/ii) × (21/two) × (21/2)

= (22/7)[(35/2) × (35/2) – (21/two) × (21/2)]

= (22/vii) × [(35² – 21^two)/iv]

= (22/7) × (56 × xiv)/4

= 616 cm²

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