The foot of the normal from the point ( 4 , 3 ) to a circle is ( 2 , 1 ) and a diameter of the circle has the equation 2 x - y - 2 = 0 , then the equation of the circle is

Question

The foot of the normal from the point ( 4 , 3 ) to a circle is ( 2 , 1 ) and a diameter of the circle has the equation 2 x - y - 2 = 0 , then the equation of the circle is, x 2 + y 2 − 4 y + 2 = 0, x 2 + y 2 − 4 y + 1 = 0, x 2 + y 2 − 2 x − 1 = 0, x 2 + y 2 − 2 x + 1 = 0

MARKS App · Accepted Answer

Equation of the diameter of the circle is given as 2 x - y - 2 = 0 ...(i) If P ( 4 , 3 ) and N ( 2 , 1 ) are the given points, then slope of PN = 3 - 1 4 - 2 = 1 Equation of normal through PN is y - 1 = ( x - 2 ) x - y - 1 = 0 ...(ii) solving (i) and (ii), we get, the centre as ( 1 , 0 ) Hence, the equation of the circle is ( x - 1 ) 2 + y 2 = ( 2 - 1 ) 2 + 1 x 2 + y 2 - 2 x - 1 = 0

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