Question

If (6, -3) is the one extremity of diameter to the circle
x2 + y 2 – 3x + 8y - 3 = 0 then its other extremity is
(b)(-3, -5)
(c) (3, -5)
(d) (3,5)​

Answer 1

Answer:

the other extermity of diameter of circle is (3, 5)

Answer 2

The other extremity of diameter is (-3,-5)

b is the correct option.

Step-by-step explanation:

The given circle is

$x^2+y^2-3x+8y-3=0$

The general equation of a circle is

$x^2+y^2+2gx+2fy+c=0$

Comparing the given equation with this equation

2 g = - 3 = > g = -3/2

2f = 8 => f = 4

Therefore, the center of the equation is given by

$(-g,-f)\\\\=(\frac{3}{2},-4)$

Now, we know that center is the mid point of the two extremities of diameter.

Let the other  extremity of diameter is (x,y).

Thus, center is the midpoint of (x,y) and (6,-3)

Formula for midpoint is given by

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

Therefore, we have

$\frac{x+6}{2}=\frac{3}{2}\\\\x+6=3\\\\x=-3$

And

$\frac{y-3}{2}=-4\\\\y-3=-8\\\\y=-5$

Therefore, the other extremity of diameter is (-3,-5)

#Learn More:

If (x,3) and (3,5) are extremities of diameter of circle with centre (2,y) then find value of x and y

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