jee/cbse 2021: circles l-2 : tangents jee/cbse 2021
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JEE/CBSE 2021: CIRCLES L-2 : Tangents
JEE/CBSE 2021: CIRCLES L-2 : Tangents
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JEE/CBSE 2021: CIRCLES L-2 : Tangents
Tangents of circle
JEE/CBSE 2021: CIRCLES L-2 : Tangents
x2 + y2 = a2
JEE/CBSE 2021: CIRCLES L-2 : Tangents
x2 + y2 + 2gx + 2fy + c = 0
JEE/CBSE 2021: CIRCLES L-2 : Tangents
Note:
1.
2.
JEE/CBSE 2021: CIRCLES L-2 : Tangents
Note:
3.
JEE/CBSE 2021: CIRCLES L-2 : Tangents
Note:
4.
JEE/CBSE 2021: CIRCLES L-2 : Tangents
Q1. Determine the equation of the tangent to the circle x2 + y2 - 2y +6x - 7 = 0 at the point F(-2, 5).
JEE/CBSE 2021: CIRCLES L-2 : Tangents
Slope Formx2 + y2 = a2
JEE/CBSE 2021: CIRCLES L-2 : Tangents
Slope Form
x2 + y2 + 2gx + 2fy + c = 0
JEE/CBSE 2021: CIRCLES L-2 : Tangents
Slope Form
JEE/CBSE 2021: CIRCLES L-2 : Tangents
Note:
1.
2.
JEE/CBSE 2021: CIRCLES L-2 : Tangents
Q2. Find equation of tangents to the circle x2 + y2 - 2x - 4y - 3 = 0 with slope 7
JEE/CBSE 2021: CIRCLES L-2 : Tangents
= 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
JEE/CBSE 2021: CIRCLES L-2 : Tangents
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
JEE/CBSE 2021: CIRCLES L-2 : Tangents
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
JEE/CBSE 2021: CIRCLES L-2 : Tangents
Tangents from a Point Outside the Circle
JEE/CBSE 2021: CIRCLES L-2 : Tangents
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
JEE/CBSE 2021: CIRCLES L-2 : Tangents
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
JEE/CBSE 2021: CIRCLES L-2 : Tangents
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
JEE/CBSE 2021: CIRCLES L-2 : Tangents
Q3. Find the common points between of line x + y = 2√2 and circle x2 + y2 = 4 .
JEE/CBSE 2021: CIRCLES L-2 : Tangents
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
JEE/CBSE 2021: CIRCLES L-2 : Tangents
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
JEE/CBSE 2021: CIRCLES L-2 : Tangents
Q4. Find the equation of the tangents to the x2 + y2 = 16 drawn from the point (1 , 4) .
JEE/CBSE 2021: CIRCLES L-2 : Tangents
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
JEE/CBSE 2021: CIRCLES L-2 : Tangents
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
JEE/CBSE 2021: CIRCLES L-2 : Tangents
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
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JEE/CBSE 2021: CIRCLES L-2 : Tangents
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)
JEE/CBSE 2021: CIRCLES L-2 : Tangents
Solution
JEE/CBSE 2021: CIRCLES L-2 : Tangents
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)
JEE/CBSE 2021: CIRCLES L-2 : Tangents
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
JEE/CBSE 2021: CIRCLES L-2 : Tangents
Solution
JEE/CBSE 2021: CIRCLES L-2 : Tangents
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
JEE/CBSE 2021: CIRCLES L-2 : Tangents
JEE/CBSE 2021: CIRCLES L-2 : Tangents
JEE/CBSE 2021: CIRCLES L-2 : Tangents
JEE/CBSE 2021: CIRCLES L-2 : Tangents
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