Question

if a line passing through the point A(0,1) and having slope 2/3 intersect thr circle x²+y²=9 at points P and Q then APxAQ
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Answer 1

AP x AQ is 8   if  a line passing through the point A(0,1) and having slope 2/3 intersect the circle x²+y²=9 at points P and Q

Given:

• A line passing through the point A(0,1)
• Slope = 2/3
• Intersect the circle x²+y²=9 at points P and Q

To Find:

• AP x AQ

Solution:

Step 1:

A line passing through the point A(0,1)  with slope 2/3

y - 1 = (2/3)(x - 0)

=> 3y - 3 = 2x

=> y = (2x + 3)/3

Step 2:

Substitute y = (2x + 3)/3 in  x²+y²=9 and solve for x

x²+{(2x + 3)/3}² =9  

=> 9x² + 4x² + 12x + 9 = 81

=> 13x² + 12x - 72 = 0

=> x=  (-12 ± √3888)/(2 * 13)

=> x = (-6 ± 18√3 )/13

Step 3:

Find y coordinate hence point P and Q

y = (2x + 3)/3

y = (11 ± 6√3)/13

P = (  (-6 +18√3 )/13 ,  (9 + 12√3)/13 )

Q =  (  (-6 -18√3 )/13 ,  (9- 12√3)/13 )

A = ( 0 , 1)

Step 4:

Find AP and AQ

AP² = ((-6 +18√3 )/13 )² + (- 4 +12√3)/13)²

=> AP² =  (- 2 + 6√3)/13)²  (3²  + 2²)

=> AP = (- 2 + 6√3)√13

AQ  = ( 2 + 6√3) /√13

Step 5:

Find the product of AP and AQ

AP.AQ

= {(- 2 + 6√3)√13  }{( 2 + 6√3) /√13}

=   104/13

= 8