If (3,-2) is the centre of a circle and 4x+3y+19=0 is a tangent to the circle, then the equation of the circle is

Question

If (3,-2) is the centre of a circle and 4x+3y+19=0 is a tangent to the circle, then the equation of the circle is (A)  x2+y2-6x+4y+25=0  (B)  x2+y2-6x+4y+12=0  (C)  x2+y2-6x+4y-12=0  (D)  x2+y2-6x+4y+13=0

Tardigrade Admin · Accepted Answer

Radius of circle = Perpendicular distance from (3,-2) to the line 4x+3y+19=0 and =(4(3)+3(-2)+19/√16+9)=5 ∴ Required equation of circle is (x-3)2+(y+2)2=52 ⇒ x2+y2-6x+4y-12=0

Q. If $(3, - 2)$ is the centre of a circle and $4 x + 3 y + 19 = 0$ is a tangent to the circle, then the equation of the circle is

$x^{2} + y^{2} - 6 x + 4 y + 25 = 0$

$x^{2} + y^{2} - 6 x + 4 y + 12 = 0$

$x^{2} + y^{2} - 6 x + 4 y - 12 = 0$

$x^{2} + y^{2} - 6 x + 4 y + 13 = 0$

$x^{2} + y^{2} - 6 x + 4 y + 9 = 0$

Radius of circle = Perpendicular distance from
$(3, - 2)$ to the line $4 x + 3 y + 19 = 0$ and
$= \frac{4 ( 3 ) + 3 ( - 2 ) + 19}{16 + 9} = 5$
$∴$ Required equation of circle is $(x - 3)^{2} + (y + 2)^{2} = 5^{2}$
$\Rightarrow$ $x^{2} + y^{2} - 6 x + 4 y - 12 = 0$

Q. If (3,−2) is the centre of a circle and 4x+3y+19=0 is a tangent to the circle, then the equation of the circle is

x2+y2−6x+4y+25=0

x2+y2−6x+4y+12=0

x2+y2−6x+4y−12=0

x2+y2−6x+4y+13=0

x2+y2−6x+4y+9=0

Radius of circle = Perpendicular distance from (3,−2) to the line 4x+3y+19=0 and =16+9​4(3)+3(−2)+19​=5 ∴ Required equation of circle is (x−3)2+(y+2)2=52 ⇒ x2+y2−6x+4y−12=0