The relationship between the zeroes and coefficients of the quadratic polynomial ax2+bx+c is
1 answer:
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Answer:
<em>The equation of the circle (x +1) )² +(y-(2))² = (2(√5))²</em>
<em> or </em>
<em>The equation of the circle x² + 2 x + y² - 4 y = 15</em>
<u>Step-by-step explanation:</u>
Given points end Points are p(-3,-2) and q( 1,6)
<em>The distance of two points formula</em>
P Q =
P Q =
P Q =
<em>The diameter 'd' = 2 r</em>
2 r = √80
=
=
<em> r = 2√5</em>
<em>Mid-point of two end points </em>
<em> </em><em></em>
<em> = (-1 ,2)</em>
<em>Mid-point of two end points = center of the circle</em>
<em> (h,k) = (-1 , 2)</em>
The equation of the circle
(x -h )² +(y-k)² = r²
<em> (x -(-1) )² +(y-(2))² = (2(√5))²</em>
<em> x² + 2 x + 1 + y² - 4 y + 4 = 20</em>
<em> x² + 2 x + y² - 4 y = 20 -5</em>
<em> x² + 2 x + y² - 4 y = 15</em>
<u><em>Final answer</em></u><em>:-</em>
<em>The equation of the circle (x +1) )² +(y-(2))² = (2(√5))²</em>
<em> or </em>
<em>The equation of the circle x² + 2 x + y² - 4 y = 15</em>
<em> </em>