A square is inscribed in a circle x^2+y^2-6x+8y-103=0
Answers
√41 is the distance of the nearest vertex to origin
Step-by-step explanation:
Complete Question is :
A square is inscribed in a circle x²+y²−6x+8y−103=0 such that its sides are parallel to co-ordinate axis
Find the distance of the nearest vertex to origin
x²+y²−6x+8y−103=0
=> (x - 3)² - 9 + (y + 4)² - 16 - 103 = 0
=> (x - 3)² + (y + 4)² = 128
=> (x - 3)² + (y + 4)² = (8√2)²
Center 3 , - 4
Radius = 8√2
=> Diagonal = 16√2
Side of Square = 16
as sides are parallel to co-odinate axis
so mid point of sides would be 8 cm away from center of circle
so Mid point of Square would be
(-5 , -4) & (11 , - 4) & (3 , -12) , (3 , 4)
Vertex would be
(-5 , -12) , (-5 , 4) & (11 , -12) , (11 , -4)
Distance of Each from origin
√5² + 12² = 13
√5² + 4² = √41
√11² + 12² = √265
√11² + 4² = √137
√41 is smallest
√41 is the distance of the nearest vertex to origin
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