Question

Find the pair of tangents drawn from (1,3) to the circle x² + y2 - 2x + 4y -11=0 and also find
the angle between them​

Answer 1

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:

xsquar -15ysquar -6x+60y-51