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Radius (OP) =\(\frac{d}{2}\) = \(\frac{26}{2}\)= 13 cm ……(i) 

∴ PR = 1/2 PQ [Perpendicular drawn from the centre of a circle to the chord bisects the chord.] 

= \(\frac{1}{2}\) x 24 = 12 cm …..(ii) 

In ∆ORP, ∠ORP = 90° 

∴ OP² = OR² + PR² [Pythagoras theorem] 

∴ 13² = OR² + 12² [From (i) and (ii)] 

∴ 169 = OR² + 144

∴ OR² = 169 – 144 

∴ OR² = 25 

∴ OR = √25 = 5 cm [Taking square root on both sides] 

∴ The distance of the chord from the centre of the circle is 5 cm.

Since, Tangent touches

A circle on a distinct point

On the diameter of a circle only two parallel tangents can be drawn.

i. Length of the arc having central angle 25° = 4 cm

Three times of 25 is 75.

Length of the arc having central angle 75° = 4 x 3 = 12 cm

ii. Length of the arc having central angle 75° and radius r = 12 cm

Length of the arc having central angle 75° and radius 1 1/2 r

\(=12\times1\frac{1}{2}=18\) cm

Correct option is: (A) diameter

The longest chord in a circle is diameter.

Correct option is: (A) diameter

Correct option is: (B) 57.75

The length of the minute hand of a clock is r = 10.5cm.

\(\because\) Area swept by minute hand in 1 hour (60 minutes).

= \(\pi r^2 \, cm^2\)

\(\therefore\) Area swept by minute hand in 10 minutes =  \(\frac {\pi r^2}{60} \times 10 cm^2\)

= \(\frac {\pi r^2}{6} = \frac 16 \times \frac {22}7 \) x 10.5 x 10.5 = 11 x 0.5 x 10.5.

= \(57.75 \, cm^2\).

Correct option is: (B) 57.75

Correct option is: (B) radius

The set of all points which are equidistant from a fixed point is a circle, then the fixed distance is called radius.

Correct option is: (B) radius

Correct option is: (C) point of contact

The point at which the tangent touches the circle is called point of contact.

Correct option is: (C) point of contact

Correct option is: (A) secant

A line, which intersects a circle in two distinct points, is called secant.

Correct option is: (A) secant

Correct option is  A) centre

Solution : True. As ∠BPA = 90°, ∠PAB = ∠OPA = 60°. Also, OP⊥PT. 

Therefore, ∠APT = 30° and ∠PTA = 60° – 30° = 30°.

Let the side of the square be x.
No. of students = x² + 24
New side = x + 1
No. of students = (x + 1)² – 25
APQ => x² + 24 = (x + 1)² – 25
=> x² + 24 = x² + 2 x + 1 - 25
=>  2x = 48
=> x = 24
Therefore, side of square = 24
No. of students = 576 + 24
= 600

Solution: 35=7+d(13-1)

➪12d=28

➪d=28/12=2.33

S13= 13/2(7+35)=273

Solution: 50=5+3(n-1)

⇒3(n-1)=45

⇒n-1=15

⇒n=16

Now, Sn=16/2(5+50)=440

Solution:37=a+3(12-1)= a+33

➪a=37-33=4

S₁₂=12/2(4+37) =246

Correct option is: (B)  \(\sqrt{119}\) cm.

\(\because\) OP \(\perp\) PQ (Radius and tangent are perpendicular at point of contact)

= \(\angle\) OPQ = 90°

Now, in right \(\triangle\) OPQ

\(PQ^2 = OQ^2 - OP^2 = 12^2 - 5^2 \)

= 144 - 25 = 119

\(\therefore\) PQ = \(\sqrt{119}\) cm

Hence, length of tangent PQ is \(\sqrt{119}\) cm.

Correct option is: (B) \(\sqrt{119}\) 

Since circle is circumscribed about the triangle whose vertices are A( – 2, 3), B(5, 2) and C(6, – 1), which implies points A, B and C are lie on circumference of circle and satisfy its equation.

The general equation of a circle: (x – h)² + (y – k) ² = r² …(i)

where (h, k) is the centre and r is the radius.

Putting A(-2, 3), B(5, 2) and C(6, -1) in above equation, we get

h² + k² + 4h – 6k + 13 = r² ……….(ii)

h² + k² – 10h – 4k + 29 = r² ………(iii)

h² + k² – 12h + 2k + 37 = r² ………(iv)

Subtract (ii) from (iii)

– 14h + 2k + 16 = 0

or – 7h + k + 8 = 0 ….(v)

Subtract (ii) from (iv)

– 16h + 8k + 24 = 0

or -2h + k + 3 = 0 ……(vi)

Solving (v) and (vi), we have

h = 1

(vi) ⇒ -2 x 1 + k + 3 = 0

⇒ k= -1

Therefore,

Centre = (1, – 1)

And,

Equation (ii) ⇒ r = 5

[using values of h and k]

Thus, required equation of the circle is

(x – 1) ² + (y + 1) ² = 5²

(x – 1) ² + (y + 1) ² = 25

Since circles are concentric, which means circles have common centre and different radii.

Equation of given circle, x² + y² + 4x + 6y + 11 = 0

The concentric circle will have the equation

x² + y² + 4x + 6y + d = 0 ….(1)

As it passes through P(5, 4),

Put x = 5 and y = 4

5² + 4² + 20 + 24 + d= 0

25 + 16 + 20 + 24 + d = 0

d = – 85

Equation (1) ⇒ x² + y² + 4x + 6y – 85 = 0

Which is required equation.

The general equation of a circle: (x – h)² + (y – k) ² = r² …(i)

where (h, k) is the centre and r is the radius.

Consider points (1, 0), (2, – 7) and (8, 1) lie on the circle.

Putting (1, 0), (2, – 7) and (8, 1) in (i)

Putting (1, 0) ⇒ h² + k² + 1 – 2h = r² ……(ii)

Putting (2, – 7) ⇒ h² + k² + 53 – 4h + 14k = r² …….(iii)

Putting (8, 1) ⇒ (8 – h)² + (1 – k) ² = r²

h² + k² + 65 – 16h – 2k = r² ………….(iv)

Subtract (ii) from (iii), we get

h – 7k – 26 = 0 ……(v)

Subtract (ii) from (iv), we get

7h + k – 32 = 0 ……(vi)

Solving (v) and (vi)

h = 5 and k = – 3

Equation (iv) ⇒ r = 25

[using h = 5 and k = – 3]

Therefore,

Centre (5, – 3)

Radius = 25

Check for (9, – 6):

To check if (9, – 6) lies on the circle,

(9 – 5)² + ( – 6 + 3)² = 5²

25 = 25

Which is true.

Hence, all the points are lie on circle.

Given r = 5, passes through points on x-axis at a distance 3 from the origin 

⇒ The points are (3,0) and (-3,0) 

Let x² + y² + 2gx + 2fy + c = 0 is required equation (3,0)6g + C + 9 = 0 … (1) 

(-3,0) -6g + C + 9 = 0 …. (2) 

r = 5; 25 = g² + f² -C 

Adding 1 and 2 we get c = -9 and g = 0 and f = ±4 

∴ The required equation is x² + y² + 8y – 9 = 0

Correct option (d)   x² + y² – 3x – 3y – 8 = 0

Explanation:

Equation of any circle passing through the point of intersection of the circles is x² + y² – 4x – 2y – 8 + λ(x² + y² – 2x – 4y – 8) = 0 This circle passes through the point (–1,4)

1 + 16 + 4 – 8 – 8 + λ(1 + 16 + 2 – 16 – 8) = 0

5 – 5λ = 0 

λ = 1

Required circle is x² + y² – 3x – 3y – 8 = 0

(a) First find the equation of the circle passing through the points 

(0,0) (1, 1) (5,-5) (6, -4) 

(0,0) ⇒ C = 0 …. (1) 

(1, 1) ⇒ 2g + 2f + 2 = 0 …. (2) 

(5,-5) ⇒ 10g -10f + 50 + c = 0 …. (3)

∴ The equation of the circle is x² + y² – 6x + 4y = 0 

Substituting the fourth point (6,-4) we get 

36 + 16 – 36 – 16 = 0 ⇒ 0 = 0 

∴ The points are concyclic. 

(b) Let us find the equation of the circle passing through (2,-4) (3,- 1) and (3, -3) we get 

(2,-4)4g – 8f + C + 20 = 0 … (1) 

(3,-1) 6g – 2f+ C + 10 = 0) … (2) 

(3,-3) 6g – 6f+c= -18 … (3) 

Solving the above equations we get g = -1, f = 2 and C = 0 

∴ Required circle is x² + y² – 2x + 4y = 0 

Substitute the fourth point (0, 0), we get 0 + 0 + 0 + 0 = 0 

∴ The four points are concyclic 

(c) Let us find the equation of the circle passing through (1, 0) (2, -7) and (8, 1) we get 

(1,0) 2g + C + 2 = 0 … (1) 

(2,-7) 4g – 14f + C + 53 = 0 … (2) 

(8,1) 16g + 2F + C + 25 = 0 … (3) 

Solving above 3 equations we get g = -\(\frac{738}{50}\),f = \(\frac{147}{50}\)and c = \(\frac{196}{25}\)

∴ Required circle is x² + y² + \(\frac{246}{50}\)x + \(\frac{147}{25}\)y + \(\frac{196}{25}\)= 0 

25x² + 25y² + 246x + 147y + 196 = 0 

Substitute the fourth point (9,-6) we get 

25(9)² + 25(-6)² + 246(9) + 147(-6) + 196 = 0 

2025 + 900 + 2214 – 882 + 196 = 0 

The points are not concyclic

Here unit circle ⇒ r = 1 

& Center of concentric circle x² + y² – 8x + 4y – 8 = 0

C = (4, -2) & r 

= \(\sqrt{(-4)^2 + (2)^2 - (C0} = \sqrt{16 + 4 - C} = 1\)

⇒ 20 – C = 19 

⇒ C = 19 

∴ The Equation of the circle is x² + y² – 8x + 4y + 19 = 0.