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Concept:

Equation of the circle having center at (h, k) and radius r is (x - h)2 + (y - k)2 = r2.

Calculation:

Let the centre of the circle on the line x + y = 2 be (a, 2 - a).

The circle will touch the x-axis and the y-axis at (a, 0) and (0, 2 - a) respectively.

Using the distance formula:

(2 - a)² = a² = r²

Now, the equation of the circle is (x - a)² + (y - 2 + a)² = r² = a².

Since, (a, 0) is a point on the circle, we must have:

0² + 4 + a² - 4a = a²

⇒ a = 1

∴ The equation of the circle is (x - 1)2 + (y - 1)2 = 12.

⇒ x² + y² - 2x - 2y + 1 = 0.

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Concept:

The center of the circle x2 + y2 + 2gx + 2fy + c = 0 is (-g, -f) and its radius r is given by r2 = g2 + f2 - c.

Calculation:

The given equation of the circle is x2 + y2 - 2x - 4y - 5 = 0.

Comparing it with the general equation of the circle we get:

g = -1, f = -2, c = -5

Using the relation r2 = g2 + f2 - c, we get:

⇒ r2 = (-1)2 + (-2)2 - (-5)

⇒ r = √10

For doubling the area, the radius must become √2 times.

∴ Required radius = √20.

Center is (-g, -f) = (1, 2).

Equation of the circle is: (x - 1)² + (y - 2)² = (√20)²

⇒ x² + y² - 2x - 4y - 15 = 0.

We have General Equation of Circle is given as 

x²+y²+2gx+2fy+c=0 ---(1)

Since, Circle passes through, (0,0) and (1,0)

Therefore Equation (1) reduces to,

c=0 ---(2)

1+2g+c=0 ---(3),

From (2) and (3), g= -1/2

Also, From As per Question, 

Cicle in equation, (1) touches, another circle having equation, 

x²+y²=(3)² ---(4)

SInce the center (0,0) of circle (4) lies on the loci of the circle (1), therefore, The radius of the circle (4) must be equal to diameter of circle(1)

hence, 2√g²+f²-c= 3

=> √1/4+f²=3/2

=> f²= 9/4-1/4

=> f²= 2

=> f= ±√2

Hence Centres of Circle (1) is (-g,-f) or (1/2,±√2)

Option (B) is correct

Concept:

The radius of the circle x² + y² + 2gx + 2fy + c = 0 is given by r² = g² + f² - c.

Calculation:

Comparing the given equation x2 + y2 - 2px + 6y - 7 = 0 with the general equation of the circle we get:

g = -p, f = 3, c = -7

Using the relation r2 = g2 + f2 - c, we get:

⇒ 52 = (-p)2 + 32 - (-7)

⇒ 25 = p2 + 9 + 7

⇒ p² = 9

⇒ p = ±3

Concept:

The equation of circle with centre at (h, k) and radius 'r' is 

(x - h)2 + (y - k)2 = r2

Calculation:

We know that, the equation of circle with centre at (h, k) and radius 'r' is 

(x - h)² + (y - k)² = r²

 Here, centre (h, k) = (2, - 3) and radius r = 5 units.

Hence, the equation of a circle with centre at (2, - 3) and radius 5 units is 

(x - 2)2 + (y + 3)2 = 52

⇒\(\rm x^2 - 4x + 4 + y^2 +6y +9 = 25\)

⇒\(\rm x^2 + y^2 - 4x + 6y - 12 = 0 \)

Hence, the equation of a circle with centre at (2, - 3) and radius 5 units is \(\rm x^2 + y^2 - 4x + 6y - 12 = 0 \).

Concept:

The general equation of a second-degree curve is 

\(\rm ax^2+2hxy+by^2+2gx+2fy+c =0\)

Where a, b, c, f, g, h are constant 

If a = b ≠ 0 and h = 0, then it represents a circle.

The general equation for the circle 

\(\rm ax^2+ay^2+2gx+2fy+c =0\)

The radius of that circle = \(\rm \sqrt{\left({g\over a}\right)^2+\left({f\over a}\right)^2-\left({c\over a}\right)}\)and center \(\rm \left(-{g\over a},-{f\over a}\right)\)

Area of a circle = πr2

Calculation: 

Given equation of the circle is:

⇒ x2 + y2 - 4x + 4y - 28 = 0

Comparing to the general equation of a circle 

\(\rm ax^2+ay^2+2gx+2fy+c =0\)

a = 1, g = -2, f = 2, c = -28

Radius = \(\rm \sqrt{\left({g\over a}\right)^2+\left({f\over a}\right)^2-\left({c\over a}\right)}\)

⇒ R = \(\rm \sqrt{\left({-2}\right)^2+\left({2}\right)^2-\left({-28}\right)}\)

⇒ R = \(\rm \sqrt{4+4+28}\)

⇒ R = \(\rm \sqrt{36}\) = 6

Diameter = 2R

D =  2 × 6 = 12 

CONCEPT :   

The distance between the centres of two circle is less than the difference of their radii then there is no common tangent.

CALCULATION: 

Equation of first circle x2 + y2   ​= 16 which can be re-written as: x2 + y2  ​= 4²

So, the centre of the first circle (0,0) and radius = 4

Equation of second circle  x2 + y2 - 2y = 0  which can be re-written as: x2 + (y-1)² = 1² 

So, the centre of the second circle is: (0,1) and radius = 1

So, the distance between the centres \(d = \sqrt {{0^2} + {1^2}} = 1\)

The difference between the radii of the two circles = |4 - 1| = 3

As we can see that, the distance between the centres < difference of their radii 

We also know that if the distance between the centres of two circle is less than the difference of their radii then there is no common tangent.

So there is no common tangent between the two given circles.

Hence, option D is the correct answer.

Concept:

Distance Formula: 

The distance 'd' between two points (x₁, y₁) and (x₂, y₂) is obtained by using the Pythagoras' Theorem:

• d² = (x₁ - x₂)² + (y₁ - y₂)²

Circle:

The equation of a circle with center (a, b) and radius r is given by:

• (x - a)² + (y - b)² = r²

Calculation:

Let's say that the points on the circle are A(-1, 1) and B(2, 1),

And the center of the circle is a point O(a, b).

Since, The center lies on the line x + 2y + 3 = 0,

The values a and b must satisfy the above linear equation

⇒ a + 2b + 3 = 0       ----(1)

Also, The distance OA = OB = radius of the circle.

Using the distance formula, we get

(a + 1)2 + (b - 1)2 = (a - 2)2 + (b - 1)2

⇒ (a + 1)² - (a - 2)² = 0

⇒ (a + 1 - a + 2)(a + 1 + a - 2) = 0

⇒ (3)(2a - 1) = 0

⇒ \(\rm a = \dfrac12\).

Using equation (1), we get:

\(\rm \dfrac12+2b+3=0\)

⇒ \(\rm b=-\dfrac{7}{4}\).

So, The center of the circle is \(\rm O\left(\dfrac{1}{2},-\dfrac{7}{4} \right)\).

Also, \(\rm r^2=OA^2=\left(\dfrac{1}{2}+1\right)^2+\left(-\dfrac{7}{4}-1\right)^2=\dfrac{9}{4}+\dfrac{121}{16}=\dfrac{157}{16}\).

Now, the equation of the circle will be:

(x - a)2 + (y - b)2 = r2

⇒ \(\rm \left (x - \dfrac12 \right )^2 + \left (y + \dfrac74 \right )^2 = \dfrac{157}{16}\)

⇒ \(\rm x^2 - x+\dfrac14 + y^2 + \dfrac72y+\dfrac{49}{16} = \dfrac{157}{16}\)

⇒ \(\rm x^2 + y^2 + \dfrac{7y}{2}- x+\left (\dfrac{4+49-157}{16} \right ) =0\)

⇒ \(\rm x^2 + y^2 + \dfrac{7y}{2}- x-\dfrac{13}{2} =0\)

∴ The equation of the circle is 2x² + 2y² - 2x + 7y - 13 = 0.

Concept:

The standard form of the equation of a circle is:

\(\rm (x-h)^2 + (y-k)^2 =R^2\)

where (h,k) are the coordinates and the R is the radius of center of the circle

Area of the circle = π R2

Note: The intersection of the Equation of diameters is center of the circle

Calculation:

Given area of circle = 154 sq.units

⇒ π R2 = 154

⇒ R2 = \(\rm 154* \frac{7}{22}\)

⇒ R = 7

Equation of the diameters

\(\rm 2x-3y+12=0\)                     ...(i)

\(\rm x+4y-5=0\)                      ...(ii)

Intersection of the diameters \(\boldsymbol{\rm (i)-2*(ii)}\)

⇒ \(\rm -11y+22=0\)

⇒ \(\boldsymbol{\rm y=2}\)

Putting back in equation (i),

⇒ \(\rm 2x-3(2)+12=0\)

⇒ \(\rm 2x=6 ⇒ \boldsymbol{\rm x=-3}\)

The center will be (-3, 2)

By the standard equation of circle

\(\rm (x-h)^2 + (y-k)^2 =R^2\)

⇒ \(\rm (x-(-3))^2+(y-2)^2 =7^2\)

⇒ \(\rm (x+3)^2+(y-2)^2 =49\)

⇒ \(\rm x^2+9+6x+y^2 +4 -4y-49=0\)

⇒ \(\boldsymbol{\rm x^2+6x+y^2-4y-36 =0}\) 

Concept:

Equatiom of circle having centre at (0, 0): \(\rm x^2+y^2=r^2\)

Let center of two circles are C₁ and C₂ and radius are r₁ and r₂.

Two circles will intersect at two points⇒ r₁ - r₂ < C₁C₂ < r₁ + r₂ 

Calculation:

Here, C₁ : \(\rm x^2+y^2=r^2\), centre of circle = (0, 0) and radius r₁ = r

And, C₂ : \(\rm x^2+y^2-6x-25=0\)⇒

\(\rm x^2-6x+9+y^2-16-9=0⇒ \rm (x-3)^2+y^2-25=0\)

⇒ \(\rm (x-3)^2+y^2=5^2\)So, centre is (3, 0) and radius, r₂ = 5

∴ C1C2 = 3 

Now two circles are intersecting so

 r1 - r2 < C1C2 < r1 + r2 

⇒ |r - 5| < 3 < r + 5

⇒ r < 8 and r + 5 > 3 ⇒ r > 2

∴ 2 < r < 8

Hence, option (4) is correct.

Concept:

The length (L) of curve y(x) in interval (a,b) is given as:

\(L = \mathop \smallint \limits_a^b \sqrt {\left[ {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} \right]dx } \)   -----(1)

Calculation:

Given:

y(x) = x^3/2

(a, b) = (0, 1)

\(\frac{{dy\left( x \right)}}{{dx}} = \frac{3}{2}\;{x^{\frac{1}{2}}}\)

From equation (1)

\(I = \mathop \smallint \limits_a^b \sqrt {\left[ {1 + \;\left( {\frac{9}{4}x} \right)} \right]} dx\)

\( = \frac{1}{2}\mathop \smallint \limits_0^1 \sqrt {\left( {4 + 9x} \right).dx} \)

Let, u = 4 + 9x

\(\frac{{\partial u}}{{\partial x}} = 9 = \frac{1}{9}\smallint du = dx\)

\(L = \left( {\frac{1}{2}} \right)\left( {\frac{1}{9}} \right)\mathop \smallint \limits_4^{13} \left( {\sqrt u } \right).du\)

\(L = \;\left( {\frac{1}{2}} \right)\left( {\frac{1}{9}} \right)\left( {\frac{2}{3}} \right){\left[ {{4^{\frac{3}{2}}}} \right]_4}^{13}\)

\(L = \frac{1}{{27}}[\left( {13{)^{\frac{3}{2}}} - {{\left( 4 \right)}^{\frac{3}{2}}}} \right]\)

\(L = \frac{1}{{27}}\left( {{{\left( {13} \right)}^{\frac{3}{2}}} - 8} \right)\)

Concept:

For circle

Area = \({\pi\over 4}× D^2\)

Circumference = π × D

The ratio of area to diameter = \({Area\over Diameter}={\pi\over 4}× D \)

where D is the diameter

Calculation:

Given:

Circumference = 264 m, \(π = {22\over 7}\)

we know that

Circumference = π × D

D = \({264\over \pi}=264×{7\over 22}\) m

D = 84 m

The ratio of area to diameter = \({Area\over Diameter}={\pi\over 4}× D \)

The ratio of area to diameter = \({{22\over 7}\over 4}×84 \)

The ratio of area to diameter = \({{22\over 7}\over 4}×84 = 66\)

Hence the ratio of area to diameter is 66 m.

Given, a circle inscribed in △ABC, such that the circle touches the sides of the triangle.  
Tangents drawn to a circle from an external point are equal.
∴AP=AR=7cm
CQ=CR=5cm
Now, BP=AB−AP=10−7=3cm
∴BP=BQ=3cm
∴BC=BQ+QC=3+5=8cm
∴ the length of BC is 8cm

Concept:

Standard equation of a circle:

\(\rm (x-h)^2+(y-k)^2=R^2\)

Where centre is (h, k) and radius is R.

Note: The normal to the circle passes through the centre of the circle.

Distance between a point on a circle and the centre is the radius of the circle.

Distance between 2 points (x1, y1) and (x2, y2) is:

D = \(\rm \sqrt{(y_2-y_1)^2 + (x_2 -x_1)^2}\)

Area of the circle = πr²

Calculation:

Given normal to the circle are :

2y - 3x + 5 = 0 ...(i)

2y + 3x - 5 = 0  ...(ii)

Substracting (ii) from (i) 

-6x + 10 = 0

⇒ x = \(\boldsymbol{5\over3}\)

From equation (ii)

2y + 3(\({5\over3}\)) - 5 = 0

⇒ y = 0

∴ Coordinates of the centre are (\(\boldsymbol{5\over3}\), 0)

Given the area of circle = 81π 

πr² = 81π 

⇒ r2 = 81

∴ The equation of circle is:

\(\rm (x-h)^2+(y-k)^2=R^2\)

⇒ (x - \({5\over3}\))2 + (y - 0)2 = 81

⇒ (x - \(\boldsymbol{5\over3}\))2 + y2 = 81

Concept:

The equation of a circle with center at O(a, b) and radius r, is given by: (x - a)² + (y - b)² = r².

Calculation:

The co-ordinates of the center of the circle are \(\rm O\left(\frac{2+0}{2},\frac{-3+1}{2}\right)\) = O(1, -1).

Radius of the circle = Distance between (0, 1) and (1, -1) = \(\rm \sqrt{(0-1)^2+(1+1)^2}\) = √5.

Equation of the circle = (x - 1)² + (y +1)² = (√5)².

⇒ x² - 2x + y² + 2y - 3 = 0.

Concept:

Standard equation of a circle:

\(\rm (x-h)^2+(y-k)^2=R^2\)

Where centre is (h, k) and radius is R.

Note: The intersection of the diameters is the centre of the circle.

Calculation:

Given diameter = 8

⇒ Radius = 4

Also diameters equations:

y + 2x - 5 = 0                    ...(i)

2y + 3x - 8 = 0                  ...(ii)

Substracting 2 × (i) from (ii)

-x + 2 = 0

x = 2

Putting back in (i)

y + 2 × 2 - 5 = 0

y = 1

So center is (2, 1) and radius 4

The equation of circle:

\(\rm (x-2)^2+(y-1)^2=4^2\)

⇒ \(\rm x^2 + 4 - 4x+y^2 + 1 - 2y = 16\)

⇒ \(\boldsymbol{\rm x^2 +y^2- 4x - 2y-11 = 0}\)

f(x, y) = 0 → circle

Let center of the circle is (-f, -g) and

f(x, y) = x² + y² + 2fx + 2gy + c = 0

Given:

f(0, a) = a² + 2ag + c = 0

roots a = 2, 3

∴ 4 + 4g + c = 0     …1)

9 + 6g + c = 0     …2)

Solving 1) and 2) we get,

9 + 6g – 4g – 4 = 0

2g = -5

g = -2.5

Now,

f(a, 0) = a² + 2fa + c = 0

roots a = 12, 0.5

∴ 144 + 24f + c = 0     …3)

0.25 + f + c = 0     …4)

Solving 3) and 4) we get,

-143.75 = 23f

f = -6.25

The center is \(\left( {\frac{{25}}{4},\;\frac{5}{2}} \right)\)