Question

2. Find the inverse point of (-2,3) with respect to the circle x² + y² - 4x – 6y +9=0
3. Find the acute angle between the circles x² + y² - 12x - 6y +41 = 0 and​

Answer 1

Answer:

n

OA is

y-2=1(x-1)

y-x=1-----(1)

then p (h,k)= p (h,h+1)

given p is inverse point of (1,2) w.rt cirde x

2

+y

2

−4x−6y+9=0

PO×OA=r

2

=(2)

2

=4

PO=

2

4

=2

2

∴(h−2)

2

+(h−2)

2

=(2

2

)

2

2(h−2)

2

=8

(h-2)=2 or h-2=-2

h=4 h=0

than k=5 or k=1

⇒p(4,5)orp(0,1)

But p and Q (1,2) lies on some side of centre of arde

⇒p=(0,1) is inverse of point of (1,2)