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If p=(-3,-2) and q=(1,6) are the endpoints of the diameter of a circle find the equation of the circle

2 Answers

2 votes

Answer:

-1 , 2 , 20

Explanation:

( x - -1 ) ² + ( y - 2 ) ² = 20

User Joebarbere
by
8.9k points
4 votes

Answer:

The equation of the circle (x +1) )² +(y-(2))² = (2(√5))²

or

The equation of the circle x² + 2 x + y² - 4 y = 15

Explanation:

Given points end Points are p(-3,-2) and q( 1,6)

The distance of two points formula

P Q =
\sqrt{x_(2) - x_(1))^(2) + ((y_(2) -y_(1))^(2) }

P Q =
\sqrt{1 - (-3)^(2) + ((6 -(-2))^(2) }

P Q =
√(16+64) = √(80)

The diameter 'd' = 2 r

2 r = √80

=
√(16 X 5)

=
4 √(5)

r = 2√5

Mid-point of two end points


((x_(1) + x_(2) )/(2) , (y_(1) +y_(2) )/(2) ) = ((-3+1)/(2) ,(-2 +6)/(2) )

= (-1 ,2)

Mid-point of two end points = center of the circle

(h,k) = (-1 , 2)

The equation of the circle

(x -h )² +(y-k)² = r²

(x -(-1) )² +(y-(2))² = (2(√5))²

x² + 2 x + 1 + y² - 4 y + 4 = 20

x² + 2 x + y² - 4 y = 20 -5

x² + 2 x + y² - 4 y = 15

Final answer:-

The equation of the circle (x +1) )² +(y-(2))² = (2(√5))²

or

The equation of the circle x² + 2 x + y² - 4 y = 15

User BevansDesign
by
8.5k points

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