JEE Advanced Model Paper 15 with solutions for online practice

P = (-4,0) Q = (0,6)
Let A = (x , y) :

PA

QA

=

2

3

⇒ 9PA2 = 4QA2
9 |(x+4)2+y2| = 4 |x2+(y−6)2|
5x2+5y2 - 72x + 48 y = 0 is equation of circum circle of ΔABC
Circumcentre of ΔABC = S = (−

36

5

,

−24

5

) and A is orthocentre
SG : GA = 1 : 2 ⇒ SG =

1

3

SA =

1

3

√(

36

5

)2+(

24

5

)2 =

4√13

5

units