The equation of a circle which has a tangent 3x + 4y = 6 and two normals given by (x - 1 ) (y - 2 ) = 0 is

Question

The equation of a circle which has a tangent 3x + 4y = 6 and two normals given by (x - 1 ) (y - 2 ) = 0 is (A)  (x - 3)2 + (y - 4)2 = 52  (B)  x2 + y2 - 4x - 2y + 4 = 0  (C)  x2 + y2 - 2x - 4y + 4 = 0  (D)  x2 + y2 - 2x - 4y + 5 = 0

Tardigrade Admin · Accepted Answer

(x-1)(y-2)=0 ⇒ x-1=0 and y-2=0 ∴ Radius =(3(1)+4(2)-6/√9+16) =(5/5)=1 ∴ Equation of the circle is (x-1)2 +(y-2)2=1 or x2+y2-2 x-4 y+4=0

Q. The equation of a circle which has a tangent $3 x + 4 y = 6$ and two normals given by $(x - 1) (y - 2) = 0$ is

$(x - 3)^{2} + (y - 4)^{2} = 5^{2}$

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

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

$x^{2} + y^{2} - 2 x - 4 y + 5 = 0$

$(x - 1) (y - 2) = 0$
$\Rightarrow x - 1 = 0$ and $y - 2 = 0$
$∴$ Radius $= \frac{3 ( 1 ) + 4 ( 2 ) - 6}{9 + 16}$
$= \frac{5}{5} = 1$
$∴$ Equation of the circle is
$(x - 1)^{2} + (y - 2)^{2} = 1$
or $x^{2} + y^{2} - 2 x - 4 y + 4 = 0$

Q. The equation of a circle which has a tangent 3x+4y=6 and two normals given by (x−1)(y−2)=0 is

(x−3)2+(y−4)2=52

x2+y2−4x−2y+4=0

x2+y2−2x−4y+4=0

x2+y2−2x−4y+5=0