Question

Find the equation of
pair of tangents from (1, 3) to
${x }^{2} + {y}^{2} - 2x + 4y - 11 = 0$
​

Answer 1

Answer:

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Step-by-step explanation:

Angle between two slopes is

tan^-1 (24/7)

•Equation of circle is :

x² + y2 - 2x + 4y -11=0

(x² +1 - 2x )+ (y²+4y+4) -11-1-4=0

•(x-1)² + (y +2)² = 16

•Centre of circle lies at (1,-2) & radius of circle is 4 units .

•Now , equation of Tangent to the circle from an external point is

(y-b) = m(x-a) ±r√(1+m²)

•where (a, b) are coordinates of center of circle ,

m is slope of Tangent

& r is radius of circle

•so , Equation of Tangent is

(y-b) = m(x-a) ±r√(1+m²)

(y+2) = m(x-1) ±4√(1+m²)

•As it passes from (1,3)

(3+2) = m(1-1) ± 4√(1+m²)

5 = ±4√(1+m²)

•squaring both sides

25 = 16 + 16m²

16m² = 25-16

m² = 9/16

m = ± 3/4

i.e. m1 = 3/4 & m2 = -3/4

•so , slope of one Tangent = 3/4

slope of other Tangent is -3/4

•let angle between them is A

tanA = (m2-m1)/(1+m1m2)

tan A = [3/4 -(-3/4)]/[1+(3/4)(-3/4)]

tanA =( 6/4)/[ 1 - 9/16]

tanA = (6/4)/(7/16)

tanA = 24/7

A = tan^-1 (24/7)

Answer 2

Answer:

$x²-y²-2x-4y+1-4x²-2x+1-y²-4y-4x²-x-x+1-{y²+4y+4}x(x-1)-1(x-1)-{y²+2y+2y+4}(x-1)(x-1)-{y(y+2)+2(y+2)}(x-1)²-{(y+2)(y+2)}$

(x-1)²-(y+2)²

(x-1-y-2)(x-1+y+2)

(x-y-3)(x+y+1)

___________ii method ________

x²+y²-2x-4y+1 = 0  --------------------(1)

2x+2yy'-2-4y' = 0                 (differentiating)

y' = (1-x)/(y-2)

since tangent is parallel to X axis , y' = 0

1-x = 0 

x = 1

put x = 1 in equation (1)

1+y²-2-4y+1 = 0

y²-4y = 0

y = 0   ,  4

hence points = (1,0)&(1,4)

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