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Question: If the centre of a circle is (2, 3) and a tangent is \(x + y = 1\) , then the equation of this circ...

If the centre of a circle is (2, 3) and a tangent is x+y=1x + y = 1 , then the equation of this circle is.

A

(x−2)2+(y−3)2=8( x - 2 ) ^ { 2 } + ( y - 3 ) ^ { 2 } = 8

B

(x−2)2+(y−3)2=3( x - 2 ) ^ { 2 } + ( y - 3 ) ^ { 2 } = 3

C

(x+2)2+(y+3)2=22( x + 2 ) ^ { 2 } + ( y + 3 ) ^ { 2 } = 2 \sqrt { 2 }

D

(x−2)2+(y−3)2=22( x - 2 ) ^ { 2 } + ( y - 3 ) ^ { 2 } = 2 \sqrt { 2 }

Answer

(x−2)2+(y−3)2=8( x - 2 ) ^ { 2 } + ( y - 3 ) ^ { 2 } = 8

Explanation

Solution

Find the radius by finding perpendicular distance from centre of the circle on the tangent and then find the equation of circle.