Question
A foot of the normal from the point (4, 3) to a circle is (2, 1), and a diameter of the circle has the equation $$2x - y = 2.$$ Then the equation of the
circle is :
A.
$${x^2} + {y^2} + 2x - 1 = 0$$
B.
$${x^2} + {y^2} - 2x - 1 = 0$$
C.
$${x^2} + {y^2} - 2y - 1 = 0$$
D.
none of these
Answer :
$${x^2} + {y^2} - 2x - 1 = 0$$
Solution :
The line joining $$\left( {4,\,3} \right)$$ and $$\left( {2,\,1} \right)$$ is also along a diameter. So, the centre is
the intersection of the diameters $$2x - y = 2$$ and $$y - 3 = \frac{{3 - 1}}{{4 - 2}} = \left( {x - 4} \right).$$
Solving these, the centre $$ = \left( {1,\,0} \right)$$
Also, the radius $$=$$ the distance between $$\left( {1,\,0} \right)$$ and $$\left( {2,\,1} \right) = \sqrt 2 .$$