Question

The point p (7,-14) lies on the circle with equation x^2 + y^2 + 6x - 14y = 483.

Answer 1

ANSWER

P has coordinates
$(-13,28)$



EXPLANATION

We need to first of all get the centre of the circle.


So let us complete the squares as follows,

${x}^{2} + {y}^{2} + 6x - 14y = 483$




${x}^{2} + 6x + {y}^{2} - 14y = 483$


${x}^{2} + 6x +{3}^{2} + {y}^{2} - 14y + ( - 7) ^{2} = 483 + {3}^{2} + {( - 7)}^{2}$



${(x + 3)}^{2} +(y - 7) ^{2} = 483 + 9+ 49$


${(x + 3)}^{2} +(y - 7) ^{2} = 541$

We compare this to the general equation of the circle

${(x - a)}^{2} + {(y - b)}^{2} = {r}^{2}$


This implies that, the center of the circle is

$C(-3,7).$




Let the coordinates of Q be
$(m,n).$

Since PQ is a diameter,the centre is the midpoint of PQ.


We now use the midpoint formula,

$(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})$


This implies that,

$C(-3,7) = ( \frac{m + 7}{2}, \frac{n - 14}{2} )$



This implies

$- 3 = \frac{m + 7}{2}$
$m + 7 = - 6$


$m = - 13$


$7 = \frac{n - 14}{2}$


$14 = n - 14$

$n = 28$

Therefore the coordinates of Q are
$(-13,28)$