If the coordinates of one end of a diameter of the circle x2+y2+4x-8y+5=0, is (2,1), the coordinates of the other end is

Question

If the coordinates of one end of a diameter of the circle x2+y2+4x-8y+5=0, is (2,1), the coordinates of the other end is (A) (-6, -7) (B) (6, 7) (C) (-6, 7) (D) (7, -6)

Tardigrade Admin · Accepted Answer

Since, one end of a diameter of the circle is

(2, 1)

.
Let the other end of a circle is

(h, k)

.
But centre of a circle is

(- 2, 4)

.

∴ \frac{h + 2}{2} = - 2

and

\frac{k + 1}{2} = 4

\Rightarrow h = - 6

and

k = 7

Hence, required point is

(- 6, 7)

.

Q. If the coordinates of one end of a diameter of the circle x2+y2+4x−8y+5=0, is (2,1), the coordinates of the other end is

(-6, -7)

(6, 7)

(-6, 7)

(7, -6)

Since, one end of a diameter of the circle is (2,1). Let the other end of a circle is (h,k). But centre of a circle is (−2,4). ∴2h+2​=−2 and 2k+1​=4 ⇒h=−6 and k=7 Hence, required point is (−6,7).

Q. If the coordinates of one end of a diameter of the circle $x^{2} + y^{2} + 4 x - 8 y + 5 = 0$ , is (2,1), the coordinates of the other end is

Since, one end of a diameter of the circle is $(2, 1)$ .
Let the other end of a circle is $(h, k)$ .
But centre of a circle is $(- 2, 4)$ .
$∴ \frac{h + 2}{2} = - 2$ and $\frac{k + 1}{2} = 4$
$\Rightarrow h = - 6$ and $k = 7$
Hence, required point is $(- 6, 7)$ .