Question

The line $3x+y-5=0$ touches a circle S at $(1,2)$. If $(h,k)$ is the ntre of the circle S such that $h^2+hk+k^2=37$ and radius is $\sqrt{10}$, then $k=$

• A. $4$
• B. $3$
• C. $2$
• D. $1$

Hint:

Tangents to a Circle.
Tangent is perpendicular to the radius at point of contact. Use point-slope form for radius and condition of perpendicularity to find relation between $h, k$.

Answer 1

The Correct Option is B

Let center be $C=(h,k)$ and point of contact $P=(1,2)$. The radius is perpendicular to the tangent. Tangent slope: $-3$; radius slope = $\frac{k - 2}{h - 1}$ \[ \frac{k - 2}{h - 1} \cdot (-3) = -1 \Rightarrow 3(k - 2) = h - 1 \Rightarrow h = 3k - 5 \quad \text{(1)} \] Radius = $\sqrt{(h - 1)^2 + (k - 2)^2} = \sqrt{10}$ \[ (h - 1)^2 + (k - 2)^2 = 10 \Rightarrow (3k - 6)^2 + (k - 2)^2 = 10 \Rightarrow 10(k - 2)^2 = 10 \Rightarrow k = 3 \text{ or } 1 \] Now plug into $h^2 + hk + k^2 = 37$: \[ \text{If } k=3,\ h = 4,\ h^2 + hk + k^2 = 16 + 12 + 9 = 37 \quad (\text{Correct}) \] \[ \text{If } k=1,\ h = -2,\ h^2 + hk + k^2 = 4 - 2 + 1 = 3 \neq 37 \quad (\text{Incorrect}) \] So $k = 3$