Question

if one end of the diameter of a circle x²+y²-4x-6y+12=0 is (1,3) . find the co-ordinate of the other end.

Answer 1

Answer:

The coordinate of other end is (3,3)

Step-by-step explanation:

equation of circle is given

$x {}^{2} + y {}^{2} - 4x - 6y + 12 = 0$

and equation of circle when the point of diameter is given

$(x - x1)(x - x2) + (y - y1)(y - y2) = 0$

and the points of diameter is

$(x1.y1) \: and(x2.y2)$

Therefore putting the value (1,3) in place if x2 and y2.

$(x - x1)(x - 1) + (y - y1)(y - 3) = 0$

$(x {}^{2} - x - x1x + x1) + (y {}^{2} - 3y - y1y + 3y1) = 0$

$((x {}^{2} + x( - 1 - x1) + x1 + y {}^{2} + y( - 3 - y1) + 3y1)) = 0$

$((x {}^{2} + y {}^{2} + ( - 1 - x1)x + ( - 3 - y1)y + (x1 + 3y1) \: )) = 0$

by comparing the given equation in question to this equestion of circle ,we get the value of other diameter point of circle.

$( - 1 - x1) = - 4$

$x1 = 3$

and

$( - 3 - y1) = - 6$

$y1 = 3$

Therefore other diameter point is (3,3). ans

Answer 2

$x² + y² - 4x - 6y + 12 = 0$

$from \: comparing \: the \: eq. \: of \: this \: Circle \: to \: the \: standard \: form \: of \: circle$

$(x² + y² + 2gx + 2fy + c = 0)$

$Here ,$

$g = -2 , f = -3 , c = 12$

$Let \: AB \: is \: the \: diameter \: of \: this \: circle , \: where \: co-ordinate \: of \: A \: is \:(1,3)$

$Centre \: of \: this \: circle -$

$C(2,3)$

$So ,$

$2 = \frac{1 + x}{2}$

$4 = x + 1$

$x = 3$

Now ,

$3 = \frac{3 + y}{2}$

$6 = 3 + y$

$y = 3$

Therefore ,

$The \: co-ordinate \: of \: other \: end \: is \: (3,3) .$