The equations of the tangents to the circle x2 + y2 - 6x + 4y = 12 , which are parallel to the straight line 4x + 3y + 5 = 0 , are

Question

The equations of the tangents to the circle x2 + y2 - 6x + 4y = 12 , which are parallel to the straight line 4x + 3y + 5 = 0 , are (A)  3x - 4y -1 9 = 0 , 3x - 4y + 31 = 0  (B)  4x + 3y - 19 = 0 , 4x + 3y + 31 = 0  (C)  4x + 3y + 19 = 0 , 4x + 3y - 31 = 0  (D)  3x - 4y + 19 = 0 , 3x - 4y + 31 = 0

Tardigrade Admin · Accepted Answer

Let equation of tangent be 4 x+3 y+k=0 centre of the circle is (3,-2) . Then, √9+4+12=|(4(3)+3(-2)+k/√16+9)| ⇒ 6+k=± 25 ⇒ k=19 text and -31 Hence, the tangent are 4 x+3 y+19=0 and 4 x+3 y-31=0

Q. The equations of the tangents to the circle $x^{2} + y^{2} - 6 x + 4 y = 12$ , which are parallel to the straight line $4 x + 3 y + 5 = 0$ , are

$3 x - 4 y - 19 = 0$ , $3 x - 4 y + 31 = 0$

$4 x + 3 y - 19 = 0$ , $4 x + 3 y + 31 = 0$

$4 x + 3 y + 19 = 0$ , $4 x + 3 y - 31 = 0$

$3 x - 4 y + 19 = 0$ , $3 x - 4 y + 31 = 0$

Let equation of tangent be $4 x + 3 y + k = 0$
centre of the circle is $(3, - 2)$ .
Then, $9 + 4 + 12 = ∣ ∣ \frac{4 ( 3 ) + 3 ( - 2 ) + k}{16 + 9} ∣ ∣$
$\Rightarrow 6 + k = \pm 25$
$\Rightarrow k = 19 and - 31$
Hence, the tangent are $4 x + 3 y + 19 = 0$ and $4 x + 3 y - 31 = 0$

Q. The equations of the tangents to the circle x2+y2−6x+4y=12 , which are parallel to the straight line 4x+3y+5=0 , are

3x−4y−19=0 , 3x−4y+31=0

4x+3y−19=0 , 4x+3y+31=0

4x+3y+19=0 , 4x+3y−31=0

3x−4y+19=0 , 3x−4y+31=0