JEE Main · 2023 · 25 Jan 2023 · Shift 2 · Circle

Points

P (- 3, 2), Q (9, 10)

and

R (α, 4)

lie on a circle

C

with PR as its diameter. The tangents to C at the points Q and R intersect at the point S . If S lies on the line

2 x - ky = 1

, then k is equal to ____ .

Answer: 3

Solution

Because

P R

is a diameter, the angle at

Q

is a right angle, so

P Q ⊥ QR

. With

m_{P Q} = \frac{10 - 2}{9 + 3} = \frac{2}{3}

, the condition

m_{P Q} m_{QR} = - 1

gives

m_{QR} = - \frac{3}{2}

, and

\frac{10 - 4}{9 - α} = - \frac{3}{2}

yields

α = 13

; thus

R = (13, 4)

and the centre is the midpoint

O = (5, 3)

. The tangent at

Q

is perpendicular to

O Q

(slope

\frac{7}{4}

), giving

4 x + 7 y = 106

, and the tangent at

R

is perpendicular to

O R

(slope

\frac{1}{8}

), giving

8 x + y = 108

. These meet at

S = (\frac{25}{2}, 8)

. Substituting into

2 x - k y = 1

gives

25 - 8 k = 1

, so

k = 3

JEE Main · 2023 · 25 Jan 2023 · Shift 2 · Integer answer

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