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If A is the midpoint of the common chord of circle x2+y24x4y=0 and x2+y2=16 and P be any point on the circumference of the circle x2+8x+y2+12x+36=0 then maximum length of AP is


Solution

First we need to find the points of intersection of two circles
Given
S1=x2+y24x4y=0      (1)
and   S2=x2+y2=16      (2)
Equation of common chord will be
S1S2=4x+4y=16     (3)
On solving eqn (2) and (3), we get intersection points are (4,0) and (0,4) and the common chord will pass from this two points
So, the mid point of the chord will be 
x=x1+x22 and y=y1+y22
The mid point A is (2,2).

Now we need to find the maximum value of AP, we know that if we draw a line segment from an external point to a circle then the line segment having maximum and minimum length length will be along the normal.

Let the centre and radius of the circle S3 be C3 and r3 where S3=x2+8x+y2+12x+36=0
Here we got C3=(4,6) and r3=4 
So the distance will be 
AP=AC3+r3
AC3=(2(4))2+(2(6))2
AC3=10

Length of AP=10+4=14
 
 

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