A circle has its diameter with endpoints at (2,3) and (8,7). Find the equation of the circle and then determine the length of a tangent drawn from the external point (10,10) to this circle.

hard 5 marks Class 9 · Mathematics · Coordinate Geometry

Answer

Equation: (x-5)^2 + (y-5)^2 = 13; Tangent length: sqrt(37)

The correct answer is Equation: (x-5)^2 + (y-5)^2 = 13; Tangent length: sqrt(37) — Compute L = sqrt(50 - 13) = sqrt(37).

Step-by-step solution

Find the midpoint of the diameter to get the centre of the circle: C((2+8)/2, (3+7)/2) = (5,5).
Compute the length of the diameter using the distance formula: d = sqrt((8-2)^2 + (7-3)^2) = sqrt(36+16) = sqrt(52).
Obtain the radius r = d/2 = sqrt(52)/2 = sqrt(13).
Write the equation of the circle: (x-5)^2 + (y-5)^2 = r^2 = 13.
Find the distance d_OC from the external point O(10,10) to the centre C(5,5): d_OC = sqrt((10-5)^2 + (10-5)^2) = sqrt(25+25) = sqrt(50).
Use the tangent length formula L = sqrt(d_OC^2 - r^2).
Compute L = sqrt(50 - 13) = sqrt(37).

Common mistake: Using radius squared instead of radius, or miscomputing distance to centre.

Concept tested: Circle radius, centre, tangent length.