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        A circle passes through the points (-1,1) , (0,6), (5,5). Find the points on the circle the tangents at which are parallel to the straight line joining origin to the centre
2 years ago

Sourabh Singh
IIT Patna
2109 Points


2 years ago
mycroft holmes
272 Points
							A(-1,1), B(0,6) and C(5,5) form a triangle right angled at B (compare slopes of AB and BC). So, the circumcentre is at mid-point of AB i.e. at (2,3), and radius = $\sqrt {13}$ i.e. origin lies on this circle. Tangent lines will therefore have the equation $3x-2y \pm 13$ (parallel to 3x-2y = 0 and at distance  $\sqrt {13}$). The intersection with the radial line 2x+3y = 0 will give you the required points as (3,-2) and (-3,2)

2 years ago
Chitrang Chauhan
11 Points
							Points are (5,1) and (-1,5). After calculating the radius (√13=3.6 units)and centre of the circle(2,3) , plot a graph on the graph paper and you will find that the tangents with the points (5,1) and (-1,5) are parallel to the line joining the origin to the centre of the circle

one year ago
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• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
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