Question

6. The circumference of the circle x2 + y2 – 18x - 16y + 120 = 0 is ​

Answer 1

Answer:

Given circle is x^2+y^2-2x-4y-20=0.

or. (x-1)^2+(y-2)^2=20+1+4

or. (x-1)^2+(y-2)^2=(5)^2

Center of the circle is A (1,2) and radius (r1)= 5 units.

Let the center of required circle is B (h,k) whose radius (r2)=5 units

Both the circles touch at point P(5,5) is given. P is mid point of AB. therefore:-

(h+1)/2=5. or. h=9

And. (k+2)/2=5. or. k=8

Required equation of circle is:-

(x-h)^2+(y-k)^2= (r2)^2

(x-9)^2+(y-8)^2=(5)^2

x^2+y^2-18x-16y+81+64=25

x^2+y^2-18x-16y+120=0. Answer.

Answer 2

Given : Circle x² + y² – 18x - 16y + 120 = 0

To Find :   circumference of the circle

Solution:

x² + y² – 18x - 16y + 120 = 0

=> (x - 9)² - 81  + (y - 8)²  -  64 + 120  = 0

=> (x - 9)²   + (y - 8)²  -25  = 0

=> (x - 9)²  + (y - 4)²  =  25

=> (x - 9)²  + (y - 4)²  =  5²

Comparing with equation of the circle

(x - h)² + (y - k)² = r²

Hence radius of the circle is 5 units

circumference of the circle  = 2πr

= 2π(5)

= 10 π units

≈ 31.4  units

circumference of the circle x² + y² – 18x - 16y + 120 = 0  is  10 π units

≈ 31.4  units

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