Let a circle C of radius 5 lie below the x-axis. The line $L_{1}: 4x+3y+2=0$ passes through the centre $P$ of the circle C and intersects the line $\mathrm{L}_{2}: 3 \mathrm{x}-4 \mathrm{y}-11=0$ at Q. The line $\mathrm{L}_{2}$ touches C at the point Q. Then the distance of P from the line $5 \mathrm{x}-12 \mathrm{y}+51=0$ is

JEE Main · 2022 · 27 Jun 2022 · Shift 2 · Circle

Let a circle C of radius 5 lie below the x-axis. The line

L_{1} : 4 x + 3 y + 2 = 0

passes through the centre

P

of the circle C and intersects the line

L_{2} : 3 x - 4 y - 11 = 0

at Q. The line

L_{2}

touches C at the point Q. Then the distance of P from the line

5 x - 12 y + 51 = 0

Answer: 11

Because

L_{2} : 3 x - 4 y - 11 = 0

is tangent to

C

Q

and

L_{1} : 4 x + 3 y + 2 = 0

runs through the centre

P

, the point

Q

is the intersection of the two lines. Solving

4 x + 3 y = - 2

together with

3 x - 4 y = 11

gives

x = 1, y = - 2

, so

Q = (1, - 2)

. The centre sits on

L_{1}

at a distance equal to the radius

5

from

Q

; a unit step along

L_{1}

\frac{( 3 , - 4 )}{5}

, so

P = (1, - 2) \pm (3, - 4)

, and the requirement that

C

lie below the

x

-axis selects

P = (4, - 6)

. The distance from

P

5 x - 12 y + 51 = 0

is then

\frac{∣5 ( 4 ) - 12 ( - 6 ) + 51∣}{5 ^{2} + 1 2 ^{2}} = \frac{∣20 + 72 + 51∣}{13} = \frac{143}{13} = 11

JEE Main · 2022 · 27 Jun 2022 · Shift 2 · Integer answer