JEE/CBSE 2021: CIRCLES L-2 : Tangents JEE/CBSE 2021 ...

JEE/CBSE 2021: CIRCLES L-2 : Tangents


Live Interactive Classes● LIVE & Interactive Teaching Style

● Fun Visualizations, Quizzes &

Leaderboards

● Daily Classes on patented WAVE platform

● Dual Teacher Model For Personal

Attention

Why Vedantu Pro is the Best?Test Series & Analysis

● Exclusive Tests To Enable Practical Application

● Mock & Subject Wise Tests, Previous Year Tests

● Result Analysis To Ensure Thorough Preparation

Assignments & Notes

● Regular Assignments To Ensure Progress

● Replays, Class Notes & Study Materials Available 24*7

● Important Textbook Solutions

● Full syllabus Structured Long Term Course

● 20+ Teachers with 5+ years of experience

● Tests & Assignments with 10,000+ questions

● Option to Learn in English or Hindi

● LIVE chapter wise course for revision

● 4000+ Hours of LIVE Online Teaching

● Crash Course and Test Series before Exam

What is Included in Vedantu Pro?

(7,500)@ ₹6000/Month

(21,000)@ ₹5,600/Month

(39,000)@ ₹5,200/Month

20% *JEE Pro Subscription Link available in description

Apply Coupon Code: NAGPRO

How to Avail The Vedantu Pro Subscription

*Vedantu Pro Subscription Link available in description

Use the code: NAGPRO

1Select Your

Grade & Target

Click on Get Subscription

2 3Apply Coupon Code and Make Payment

Choose Your Subscription & Click on Proceed To Pay

4


Tangents of circle


x2 + y2 = a2


x2 + y2 + 2gx + 2fy + c = 0


Note:

1.

2.


Note:

3.


Note:

4.


Q1. Determine the equation of the tangent to the circle x2 + y2 - 2y +6x - 7 = 0 at the point F(-2, 5).


Slope Formx2 + y2 = a2


Slope Form

x2 + y2 + 2gx + 2fy + c = 0


Slope Form


Note:

1.

2.


Q2. Find equation of tangents to the circle x2 + y2 - 2x - 4y - 3 = 0 with slope 7


= 0x2 + y2 – 2x – 4y – 32g = – 2, g = – 12f = – 4, f = – 2, c = – 3

∴ Centre ≡ (–g, –f ) ≡ (1,2) Radius r

==

=

√g2 + f2 – c √1 + 4 –(–3) √8

∴

∴

Comparing with

x2 + y2 + 2gx + 2fy + c = 0

Solution


y = 7x + c

7x – y + c = 0

Let p is ⊥ distance between centre (1, 2) & 7x – y + c=0

Equation of tangent is, y =m x + c

p =7(1) – 2 + c

√72 + (–1)2

p = 5 + c √50

1

Slope = m = 7

Solution


5 + c =

5 + c = ± 20c = 20 – 5 or c = –20 – 5c = 15 or c = – 25∴ Equations of tangents,

7x – y + 15 = 0 or 7x – y – 25 = 0

± √8 .√50

p = 5 + c √50

= √8r

∴ Line is tangent

p = r

5 + c √50

√8=

Solution


Tangents from a Point Outside the Circle


Tangents

a) At point (x1 , y1)

b) At parametric point

Standard circle

x2 + y2 = a2 x2 + y2 + 2gx + 2fy + c = 0

General circle

xx1 + yy1 = a2 xx1 + yy1 + g(x + x1) + f(y + y1) +

x cos θ + y sin θ = a N. A.(acos θ , asinθ)

c = 0


c) Slope form y = mx ± a√1 + m2 i) y = mx + cii) To find c :- Use

perpendicular distance

between centre and

tangent is equal to radius .

Tangents

Standard circle

x2 + y2 = a2 x2 + y2 + 2gx + 2fy + c = 0

General circle


d) From point (x1 , y1)

i) y = mx ± a√1 + m2 i) y – y1 = m(x – x1)ii) To find m :- Use

perpendicular distance between centre and tangent is equal to radius .

ii) y1 = mx1 ± a√1 + m2

solve for values of m

iii) Use: y – y1 = m(x – x1)

Tangents

Standard circle

x2 + y2 = a2 x2 + y2 + 2gx + 2fy + c = 0

General circle


Q3. Find the common points between of line x + y = 2√2 and circle x2 + y2 = 4 .


x2 + y2 = 4 . . . .(i)x + y = 2√2

y = 2√2 – x . . . .(ii)

solving (i) and (ii) ,

x2 + (2√2 – x)2 = 4x2 + 8 – 4√2x + x2 = 42x2 – 4√2x + 4 = 0

To find common points solve the given equation simultaneously

Solution


2x2 – 4√2x + 4 = 0Dividing by 2 ,x2 – 2√2x + 2 = 0

∴ (x –√2 )2 = 0

∴ x = √2

Substituting x = √2 in equation (ii) ,

y = 2√2 – √2∴ y = √2∴ Common point is (√2 , √2 )∴ Line is tangent to the circle .

y = 2√2 – x …(ii)

Solution


Q4. Find the equation of the tangents to the x2 + y2 = 16 drawn from the point (1 , 4) .


Given circle is x2 + y2 = 16

Equation of tangent is ,

y = mx + a√1 + m2Slope form ,

As tangent passes through (1 , 4)

y = mx + 4√1 + m2

4 = m + 4√1 + m2

(4 – m)2 = 16 (1 + m2)

Solution


16 – 8m + m2 = 16 + 16m2

15m2 + 8m = 0

m(15m + 8) = 0

m = 0 or 15m + 8 = 0

m = 0 or m = –815

Equation of tangent is ,y – y1 = m(x – x1)

If m = 0 and (x1 , y1) ≡ (1 , 4)

(4 – m)2 = 16 (1 + m2)

Solution


y – 4 = 0

If m = –815

y – 4 =–815

(x – 1)

and (x1 , y1) ≡ (1 , 4)

15y – 60 = –8x + 8

8x + 15y – 68 = 0

Solution

*JEE Pro Subscription Link available in description

Apply Coupon Code: NAGPRO

(5,000)@ ₹4000/Month

(13,500)@ ₹3600/Month

(24,000)@ ₹3200/Month

(42,000)@ ₹2800/Month

20%


Q5. Equation of the tangent to the circle, at the point (1, -1), whose centre is the point of intersection of the straight lines x - y - 1 and 2x + y - 3 is :

A

B

4x + y - 3 =0

x + 4y + 3 =0

D

C

x - 3y - 4 = 0

3x - y - 4 =0

A

B

D

C

D

Jee Main 2016 (Online)


Solution


Q5. Equation of the tangent to the circle, at the point (1, -1), whose centre is the point of intersection of the straight lines x - y - 1 and 2x + y - 3 is :

A

B

4x + y - 3 =0

x + 4y + 3 =0

D

C

x - 3y - 4 = 0

3x - y - 4 =0

A

B

D

C

D

Jee Main 2016 (Online)


Q6. Tangent to circle x2 + y2 = 5 at (1, -2) also touches the circle x2 + y2 - 8x + 6y + 20 =0. Find the coordinate of the corresponding point of contact

A

B

(3, - 2)

(3, 1)

D

C

(3, 2)

(3, -1)

A

B

D

C

D


Solution


Q6. Tangent to circle x2 + y2 = 5 at (1, -2) also touches the circle x2 + y2 - 8x + 6y + 20 =0. Find the coordinate of the corresponding point of contact

A

B

(3, - 2)

(3, 1)

D

C

(3, 2)

(3, -1)

A

B

D

C

D


Thank You

JEE/CBSE 2021: CIRCLES L-2 : Tangents JEE/CBSE 2021 ...

Documents

Transcript of JEE/CBSE 2021: CIRCLES L-2 : Tangents JEE/CBSE 2021 ...