Manipal Entrance Test 2015 Math Paper

Since, 3 x + y = 0 is a tangent to the circle with centre at (2, - 1).
∴ Radius = Length of the perpendicular from (2, -1 ) on 3 x + y = 0
⇒ Radius =

6−1

√9+1

=

5

√10

=√

5

2

So, the equation of the circle is
(x−2)2+(y+1)2=

5

2

⇒ x2+y2−4x+2y+

5

2

=0
The combined equation of the tangents drawn from the origin to this circle is
SS1=T2
where, S=x2+y2−4x+2y+

5

2

,
S1=02+02−4×0+2×0+

5

2

=

5

2

and T=x(0)+y(0)−2(x+0)+(y+0) +

5

2

=−2x+y+

5

2

∴ (x2+y2−4x+2y+

5

2

)(

5

2

) =(−2x+y+

5

2

)2
⇒ 3x2−8xy−3y2=0
⇒ 3x+y=0 and x−3y=0