Q3 of 112 Page 146

The center of a circle is at (–6, 4). If one end of a diameter of the circle is at the origin, then find the other end.

Here, one end of the diameter is at origin that is A (0,0) and let B (a, b) is the required endpoint of the diameter.


Center {O (–6, 4)} of the circle will be exactly middle of the diameter. So, to find another we have to use the mid–point formula


Midpoint of line–segment joining the points (x1, y1) and (x2, y2)


M (x, y) = M


⇒ (–6, 4) =


⇒ (–6, 4) =


and


a = –6 × 2 and b = 4 × 2


a = –12 and b = 8


Therefore, another end point of the diameter (–12, 8)


More from this chapter

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1

Find the midpoint of the line segment joining the points

(i) (1, – 1) and (–5, 3) (ii) (0,0) and (0,4)

 2

Find the centroid of the triangle whose vertices are

(i) (1,3), (2, 7) and (12, – 16)


(ii) (3, – 5), (–7, 4) and (10, – 2)

 4

If the centroid of a triangle is at (1, 3) and two of its vertices are (–7, 6) and (8, 5) then find the third vertex of the triangle.

 5

Using the section formula, show that the points A (1,0), B (5,3), C (2,7) and D (–2, 4) are the vertices of a parallelogram taken in order.