Question

The length of the tangents from (1,1)to the circle 2x^2+2y^2+5x+3y+1=0 is

Answer 1

we have to find the length of the tangents which are drawn from (1,1) to the circle 2x² + 2y² + 5x + 3y + 1 = 0

solution : equation of circle ; 2x² + 2y² + 5x + 3y + 1 = 0

⇒x² + y² + (5/2)x + (3/2)y + (1/2) = 0

centre ; O = (-5/4, -3/4)

radius, r = √{(5/4)² + (3/4)² - 1/2}

= √{25/16 + 9/16 - 8/16)

= √(13/8)

let a tangent touches the circle at T , tangent drawn from the point P(1,1) and Centre O (-5/4, -3/4)

here, OP² = TP² + TO²

OP = √{(-5/4 - 1)² + (-3/4 - 1)²}

= √{81/16 + 49/16}

= √(130/16)

OP² = 130/16 = 65/8

TO = radius = √(13/8)

so, TO² = 13/8

now 65/8 = TP² + 13/8

⇒52/8 = TP²

⇒TP² = 13/2

⇒TP = √(13/2)

Therefore the length of tangent is √(13/2)