If m1 and m2 are the slopes of tangents to the circle x 2 + y 2 = 4 from the point (3, 2) , then m1 - m2 is equal to

Question

If m1 and m2 are the slopes of tangents to the circle x 2 + y 2 = 4 from the point (3, 2) , then m1 - m2 is equal to (A)  (5/12)  (B)  (12/5)  (C)  (3/2)  (D)  0

Tardigrade Admin · Accepted Answer

Equation of pair of tangents is S S1=T2 ⇒ (x2+y2-4)(9+4-4)=(3 x+2 y-4)2 ⇒ 5 y2+16 y-12 x y+24 x-52=0 ∴ m1+m2=(-2 h/b)=(12/5) and m1 m2=0 Now, m1-m2=√(m1+m2)2-4 m1 m2 =√((12/5))2-0 =(12/5)

Q. If $m_{1}$ and $m_{2}$ are the slopes of tangents to the circle $x^{2} + y^{2} = 4$ from the point $(3, 2)$ , then $m_{1} - m_{2}$ is equal to

$\frac{5}{12}$

$\frac{12}{5}$

$\frac{3}{2}$

$0$

Q. If m1​ and m2​ are the slopes of tangents to the circle x2+y2=4 from the point (3,2) , then m1​−m2​ is equal to

125​

512​

23​

0

Equation of pair of tangents is SS1​=T2 ⇒(x2+y2−4)(9+4−4)=(3x+2y−4)2 ⇒5y2+16y−12xy+24x−52=0 ∴m1​+m2​=b−2h​=512​ and m1​m2​=0 Now, m1​−m2​=(m1​+m2​)2−4m1​m2​​ =(512​)2−0​ =512​

Q. If $m_{1}$ and $m_{2}$ are the slopes of tangents to the circle $x^{2} + y^{2} = 4$ from the point $(3, 2)$ , then $m_{1} - m_{2}$ is equal to

$\frac{5}{12}$

$\frac{12}{5}$

$\frac{3}{2}$

$0$