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 2x-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 2x-y-2=0 , then the equation of the circle is (A) x2 + y2 - 4 y + 2 = 0 (B) x2 + y2 - 4 y + 1 = 0 (C) x2 + y2 - 2 x - 1 = 0 (D) x2 + y2 - 2 x + 1 = 0

Tardigrade Admin · 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 = \frac{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

Q. 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

$x^{2} + y^{2} - 4 y + 2 = 0$

$x^{2} + y^{2} - 4 y + 1 = 0$

$x^{2} + y^{2} - 2 x - 1 = 0$

$x^{2} + y^{2} - 2 x + 1 = 0$

Q. 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 2x−y−2=0 , then the equation of the circle is

x2+y2−4y+2=0

x2+y2−4y+1=0

x2+y2−2x−1=0

x2+y2−2x+1=0

Q. 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

$x^{2} + y^{2} - 4 y + 2 = 0$

$x^{2} + y^{2} - 4 y + 1 = 0$

$x^{2} + y^{2} - 2 x - 1 = 0$

$x^{2} + y^{2} - 2 x + 1 = 0$