Question

The length of common tangents to circles x2+ y2 - 6x - 4y +9=0 & x² + y = 1 is​

Answer 1

Given : circles x2+ y2 - 6x - 4y +9=0 & x² + y ²= 1

To Find :  length of common tangents to circles

Solution:

x²+ y² - 6x - 4y +9=0

( x- 3)² + ( y - 2)² = 2²

center = ( 3 , 2)   , r= 2

x² + y² = 1

Center = (0,0) , r = 1

Distance between center =  √13

Let say point is at distance x from 0,0 from where common tangents are drawn .

a/(a + √13) = 1/2

=> 2a = a + √13

=> a = √13

Length of tangent to small circle

= √13 - 1  = √12  = 2√3

Length of tangent to Larger circle

= √52 - 4  = 4√3

Length of tangent between circles = 2√3

tangent between circle intersection point at distance x from origin

x / √13 - x  =  1/2

=> 2x = √13 - x

=> x = √13/3

Length of Tangent = √(√13/3)² - 1²   + √(2√13/3)² - 2²

= (1/3)√4 + (2/3)√4

=  2

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Answer 2

The equation of the common chord is

S 2 −S 1 =0(x 2 +y 2 −6x−4y+9)−(x 2 +y −8x−6y+23)=2x+2y−14=0

x+y=7...(i)

Now this is the diameter of x

2 +y 2 −8x−6y+23=0

Hence it should cut the circle at two distinct points.Therefore x 2 +(7−x) 2 −8x−

2x 2

+x(−14−8+6)+49−42+23=02x 2

−16x+30=0x 2 −8x+15=0

x=5,3y=2,4 repectively.D= (x 1 −x 2 ) 2 +(y 1 −y 2 ) 2= 2 2 +2 2

=2√ 2 .