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

Two circles are said to congruent if and only if they have the same radii.

Circles having the same centre are called concentric circles.

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

A line segment joining any two points on a circle is called the chord of the circle.

(a-b)x² + (b-c) x+ (c - a) = 0
T.P 2a = b + c
B² – 4AC = 0
(b-c)² – [4(a-b) (c - a)] = 0
b²-2bc + c² – [4(ac-a² – bc + ab)] = 0
 b2-2bc + c² – 4ac + 4a² + 4bc - 4ab = 0
 b²+ 2bc + c² + 4a² – 4ac – 4ab= 0
 (b + c - 2a)² = 0
 b + c = 2a

i. PQRS is a cyclic quadrilateral. [Given] 

∴ ∠PSR + ∠PQR = 180° [Opposite angles of a cyclic quadrilateral are supplementary]

∴ 110° + ∠PQR = 180° 

∴ ∠PQR = 180° – 110° 

∴ m ∠PQR = 70° 

ii. ∠PSR= 1/2 m (arcPQR) [Inscribed angle theorem] 

110°= 1/2 m (arcPQR) 

∴ m(arc PQR) = 220° 

iii. In ∆PQR, 

side PQ ≅ side RQ [Given] 

∴ ∠PRQ = ∠QPR [Isosceles triangle theorem] 

Let ∠PRQ = ∠QPR = x

Now, ∠PQR + ∠QPR + ∠PRQ = 180° [Sum of the measures of angles of a triangle is 180°] 

∴ ∠PQR + x + x= 180° 

∴ 70° + 2x = 180° 

∴ 2x = 180° – 70° 

∴ 2x = 110°

∴ x = 100°/2 = 55°

∴ ∠PRQ = ∠QPR = 55°….. (i) 

But, ∠QPR = 1/2 nm(arc QR) [Inscribed angle theorem] 

∴ 55° = 1/2 m(arc QR) 

∴ m(arc QR) = 110° 

iv. ∠PRQ = ∠QPR =55° [From (i)] 

∴ m ∠PRQ = 55° 

The general form of the equation of a circle is:

(x – h)² + (y – k)² = r² …..(1)

Where, r is the radius of the circle and (h, k) is the centre of the circle.

Here, r = 6, h = -3 and k = -2

Equation (1)

⇒ (x + 3)² + (y + 2)² = 6²

or (x + 3)² + (y + 2)² = 36

or x² + y² + 6x + 4y – 23 = 0

Which is the required equation.

The general form of the equation of a circle is:

(x – h)² + (y – k)² = r² …..(1)

Where, r is the radius of the circle and (h, k) is the centre of the circle.

Here, r = √2, h = a and k = a

Equation (1)⇒

(x – a)² + (y – a)² = (√2)²

or (x – a)² + (y – a)² = 2

or x² + y² – 2ax – 2ay + (2a² – 2) = 0

Which is the required equation.

The general form of the equation of a circle is:

(x – h)² + (y – k)² = r² …..(1)

Where, r is the radius of the circle and (h, k) is the centre of the circle.

Here, r = √(a² – b²), h = -a and k = -b

Equation (1)⇒

(x + a )² + (y + b)² = (√(a² – b²))²

or (x + a )² + (y + b)² = a² – b²

or x² + y² + 2a x + 2ay + a² + b² = a² – b²

or x² + y² + 2a x + 2ay + 2 b² = 0

Which is the required equation.

The general form of the equation of a circle is:

(x – h)² + (y – k)² = r² …..(1)

Where, r is the radius of the circle and (h, k) is the centre of the circle.

Here, r = 4, h = 0 and k = 0

Equation (1)⇒

(x – 0 )² + (y – 0)² = (4)²

or x² + y² = 16

or x² + y² – 16 = 0

Which is the required equation.

The general form of the equation of a circle is:

(x – h)² + (y – k) ² = r² …..(1)

Where, (h, k) is the centre of the circle.

r = radius of the circle.

We are given with, centre = (2, – 5)

Or (h, k) = (2, – 5)

Find the radius of circle:

Since the circle passes through (3, 2), so it must satisfy the equation.

Put x = 3 and y = 2 in (1)

(3 – 2)² + (2 + 5) ² = r²

1 + 49 = r²

Or r² = 50

Now,

Equation of circle is:

(x – 2)² + (y + 5)² = 50

Which is required equation.

Given x² + y² – 4x + 2y – 1 = 0 

P = (2,-1) & centre of the circle is point of intersection of diameters 

x + y = 3 – (1) & 2x + y = 2 – (2) 

eqn. (2) – eqn. (1) gives x = -1 & y = 3 – x = (3 – C – 1) = 4 

∴ centre = c(-1, 4) 

r = CP 

=  \(\sqrt{(-1 - 2)^2 - (4 - (-1)^2} - \sqrt{(-3)^2 + 5^2} = \sqrt{9 + 25}\)

= √34

∴ the eqn. of the circle with c = (-1, 4) & r = √34 is 

(x + 1)² + (y – 4)² = (√34)²

x² + 1 + 2x + y² + 16 – 8y = 34 

x² + y² + 2x – 8y – 17 = 0.

Correct option is: (C) 1

(a) x² + y² – 4x – y – 5 = 0 

Comparing with x² + y² + 2gx + 2fy + C = 0 

We get 2g = – 4, 2f = – 1 ⇒ g = -2, f = \(\frac{1}{2}\)

∴ C=(-8,-1) = (2, \(\frac{1}{2}\)) 

(b) Given 3x² + 3y² – 6x – 12y – 2 = 0, divide by 3 

x² + y² – 2x – 4y – \(\frac{2}{3}\) = 0 

Here g = -1, f = -2, & c = \(\frac{2}{3}\)

∴Centre = (-g, -f) = (1, 2) & r = \(\sqrt{g^2 + f^2 - c}\)

= \(\sqrt{(-1)^2 + (-2)^2 - (-\frac{2}{3}})\)

r = \(\sqrt{1 - 4 - \frac{2}{3}}\) = \(\sqrt{\frac{17}{3}}\) units.

(c) Given (x – 2) (x – 4) + (y – 1) (y – 3) = 0 

⇒ x² – 6x + 8 + y² – 4y + 3 = 0 

⇒ x² + y² – 6x – 4y + 11 = 0 

Here g = -3, f= -2, C = 11 & 

∴ Centre = (3, 2).

And r = \(\sqrt{9 + 4 - 11}\) = √2 units

(d) Given x² + y² – 2xcosα – 2ysinα – 1 = 0 

g = – cosα, f = – sinα, c = -1 

∴ Centre = (cosα, sinα) & r 

= \(\sqrt{cos^\alpha + sin^2\alpha + 1} = \sqrt{1 + 1}\) = √2

Given x² + y² + 4x – 2y – k = 0 & r = 4 units, K = ? 

Here g = 2, f = -1 and C = – k 

∴ Centre = (-2, 1) & r = \(\sqrt{g^2 + f^2 - c}\)

4 = \(\sqrt{(2)^2 + (-1)^2 - (-k)}\) ⇒ 4 = \(\sqrt{4 + 1 + k}\) S.B.S

16 = 5 + k 

⇒ k = 16 – 5 = 11

Given (x_1, y₁) = (a, b) (x₂, y₂) = (-5, 1) & Center of the given circle = (-2, 2). 

Equation of the circle when ends of diameter is given is (x – x₁) (x – x₂) + (y – y₁) (y – y₂) = 0. 

(x – a) (x + 5) + (y – b) (y – 1) = 0. 

⇒ x² – 5a – ax + 5x + y² – by – y + b = 0. 

x² + y² + x (5 – a) – y(b + 1) + (b – 5a) = 0 comparing. 

This by given circle x² + y³ + 4x – 4y – 2 = 0 

we get 5 – a = 4, b + 1 = 4 

⇒ 5 – 4 = a, b = 4 – 1 

∴ a = 1 & b = 3.

Given x² + y² – 4x – 8y + k = 0 is a point circle 

⇒ r = 0, g = -2, f = -4 & c = k

and \(\sqrt{g^2 + f^2 - C} = r\)

= \(\sqrt{4 + 16 - k} = 0\)

\(= \sqrt{20 - k} = 0\)

⇒ k = 20 S.B.S

To Prove: ∠APB + ∠AOB = 180° 

OA is radius, PA is tangent. 

∴∠PAO = 90° OB is radius, PB is tangent. 

∴ ∠PBO = 90° 

Now OAPB is a quadrilateral. 

∴∠PAO + ∠PBO = 90° + 90° = 180° 

Sum of four angles of a quadrilateral is 360° 

∴ ∠PAO + ∠PBO + ∠APB + ∠AOB = 360° 

180° + ∠APB + ∠AOB = 360° 

∠APB + ∠AOB = 360° – 180° 

∴ ∠APB + ∠AOB = 180° 

If sum of two angles is equal to 180°, then they are supplementary angles. 

∴ ∠APB and ∠AOB are supplementary angles.

Data: In the quadrilateral, ABCD is drawn to circumscribe a circle. 

To Prove: AB + CD = AD + BC 

The lengths of tangents drawn from an extrenal point to a circle are equal. 

∴ Tangents AP and AS are drawn from point A. 

∴ AP = AS Tangents BP and BQ are drawn from point B. 

∴ BP = BQ Tangents CQ, CR are drawn from point C. 

∴ CQ = CR. Tangents DR, DS are drawn from point A. 

∴ DR = DS 

L.H.S.: AB + CD = AP + PB + CR + RD 

= AS + BQ + CQ + DS 

= AS + SD + BQ + CQ 

∴ AB + CD = AD + BC 

∴ L.H.S. = R.H.S.

Data: XY and X’Y’ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X’Y’ at B 

To Prove: ∠AOB = 90°. 

Tangent XY || Tangent X’Y’ 

∴ ∠PAB + ∠QBA = 180° (∵ interior angles) 

\(\frac{1}{2}\)∠PAB + \(\frac{1}{2}\)∠QBA = \(\frac{1}{2}\) x 180°

∠OAB + ∠OBA = 90° 

Now, in AOAB, 

∠AOB + ∠OAB + ∠OBA = 180° 

∠AOB + 90° = 180° 

∴ ∠AOB = 180° – 90° 

∴ ∠AOB = 90°.

i. Here, circle with centres A and B touch each other externally at point C. 

∴ d(A, B) = d(A, C) + d(B ,C)

= 3 + 4 

∴ d(A,B) = 7 cm 

[The distance between the centres of circles touching externally is equal to the sum of their radii] 

ii. Here, circle with centres A and 13 touch each other internally at point C. 

∴ d(A, B) = d(A, C) – d(B, C) 

= 4 – 3 

∴ d(A,B) = 1 cm

[ The distance between the centres of circles touching internaly is equal to the difference in their radii]

The circles with centres P and Q touch each other at R. 

∴ By theorem of touching circles, 

P – R – Q 

i. In ∆PAR, seg PA = seg PR [Radii of the same circle]

∴ ∠PRA ≅ ∠PAR (i) [Isosceles triangle theorem] 

Similarly, in ∆QBR, 

seg QR = seg QB [Radii of the same circle] 

∴ ∠RBQ ≅ ∠QRB (ii) [Isosceles triangle theorem] 

But, ∠PRA ≅ ∠QRB (iii) [Vertically opposite angles] 

∴ ∠PAR ≅ ∠RBQ (iv) [From (i) and (ii)] 

But, they are a pair of alternate angles formed by transversal AB on seg AP and seg BQ.

∴ seg AP || seg BQ [Alternate angles test] 

ii. In ∆APR and ∆RQB, 

∠PAR ≅ ∠QRB [From (i) and (iii)] ∠APR ≅ ∠RQB [Alternate angles]

∴ ∆APR – ∆RQB [AA test of similarity] 

iii. ∠PAR = 35° [Given] 

∴ ∠RBQ = ∠PAR= 35° [From (iv)] 

In ∆RQB, 

∠RQB + ∠RBQ + ∠QRB = 180° [Sum of the measures of angles of a triangle is 180°] 

∴ ∠RQB + ∠RBQ + ∠RBQ = 180° [From (ii)] 

∴ ∠RQB + 2 ∠RBQ = 180° 

∴ ∠RQB + 2 × 35° = 180° 

∴ ∠RQB + 70° = 180° 

∴ ∠RQB = 110°

Circles with centers N and M lie in the same plane and intersect with a line p in the plane in one and only one point T [K – N – M].

Hence, the given circles are internally touching circles.

i. ∠ACB = 2 ∠ADB. 

ii. ∠ADB + ∠AEB = 180°. 

iii. ∠AHB = ∠ADB = ∠AFB = ∠AGB 

iv. ∠AEB = ∠AIB

∠PRQ = ∠PSQ = ∠PTQ = 90° 

Circles with centres R and S lie in the same plane and intersect with a line l in the plane in one and only one point T [R – T – S].

Hence the given circles are externally touching circles.