Find the equation of circle having normals (x-1)(y-2)=0 and a tangent 3 x+4 y=6 ?

Question

Find the equation of circle having normals (x-1)(y-2)=0 and a tangent 3 x+4 y=6 ? (A) (x-1)2+(y-2)2=1 (B) (x-2)2+(y-1)2=1 (C) (x+1)2+(y+2)2=1 (D) (x+2)2+(y+1)2=1

Tardigrade Admin · Accepted Answer

The equation of normals to the circle are

x - 1 = 0

and

y - 2 = 0

, so centre of the circle is

(1, 2)

and since

3 x + 4 y = 6

is the tangent to the circle so radius

r = \frac{3 + 8 - 6}{3 ^{2} + 4 ^{2}} = 1

∴

Equation of required circle is

(x - 1)^{2} + (y - 2)^{2} = 1

Q. Find the equation of circle having normals $(x - 1) (y - 2) = 0$ and a tangent $3 x + 4 y = 6$ ?

$(x - 1)^{2} + (y - 2)^{2} = 1$

$(x - 2)^{2} + (y - 1)^{2} = 1$

$(x + 1)^{2} + (y + 2)^{2} = 1$

$(x + 2)^{2} + (y + 1)^{2} = 1$

The equation of normals to the circle are $x - 1 = 0$ and $y - 2 = 0$ , so centre of the circle is $(1, 2)$ and since $3 x + 4 y = 6$ is the tangent to the circle so radius $r = \frac{3 + 8 - 6}{3 ^{2} + 4 ^{2}} = 1$
$∴$ Equation of required circle is
$(x - 1)^{2} + (y - 2)^{2} = 1$

Q. Find the equation of circle having normals (x−1)(y−2)=0 and a tangent 3x+4y=6 ?

(x−1)2+(y−2)2=1

(x−2)2+(y−1)2=1

(x+1)2+(y+2)2=1

(x+2)2+(y+1)2=1