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 $k = 3$

Answer 2

Step 1: Understand the problem
The line \(3x + y - 5 = 0\) touches the circle \(S\) at the point \((1, 2)\). The center of the circle is \((h, k)\), and it satisfies:
\[ h^2 + hk + k^2 = 37 \] The radius of the circle is \(\sqrt{10}\). We need to find the value of \(k\).

Step 2: Use the tangent condition
Since the line touches the circle at \((1, 2)\), the point lies on both the circle and the line.
The center \((h, k)\) lies such that the radius is perpendicular to the tangent at the point of contact.
The radius vector is \(\overrightarrow{CP} = (1 - h, 2 - k)\), and the tangent line has normal vector \(\vec{n} = (3, 1)\).
Since radius is perpendicular to the tangent, \(\overrightarrow{CP}\) is parallel to \(\vec{n}\), so:
\[ \frac{1 - h}{3} = \frac{2 - k}{1} \] which simplifies to: \[ 1 - h = 3(2 - k) \implies 1 - h = 6 - 3k \implies h = 1 - 6 + 3k = -5 + 3k \]

Step 3: Use the radius length condition
The radius length is: \[ \sqrt{(1 - h)^2 + (2 - k)^2} = \sqrt{10} \] Square both sides: \[ (1 - h)^2 + (2 - k)^2 = 10 \] Substitute \(h = -5 + 3k\): \[ (1 - (-5 + 3k))^2 + (2 - k)^2 = 10 \] \[ (1 + 5 - 3k)^2 + (2 - k)^2 = 10 \] \[ (6 - 3k)^2 + (2 - k)^2 = 10 \] \[ (6 - 3k)^2 = 36 - 36k + 9k^2 \] \[ (2 - k)^2 = 4 - 4k + k^2 \] Sum: \[ 36 - 36k + 9k^2 + 4 - 4k + k^2 = 10 \] \[ 40 - 40k + 10k^2 = 10 \] \[ 10k^2 - 40k + 40 = 10 \] \[ 10k^2 - 40k + 30 = 0 \] Divide both sides by 10: \[ k^2 - 4k + 3 = 0 \] Factor: \[ (k - 3)(k - 1) = 0 \] So, \[ k = 3 \quad \text{or} \quad k = 1 \]

Step 4: Use the given equation \(h^2 + hk + k^2 = 37\) to find the correct \(k\)
Recall \(h = -5 + 3k\), substitute and check both values:
For \(k=3\): \[ h = -5 + 3(3) = -5 + 9 = 4 \] Calculate: \[ h^2 + hk + k^2 = 4^2 + 4 \times 3 + 3^2 = 16 + 12 + 9 = 37 \] Correct.
For \(k=1\): \[ h = -5 + 3(1) = -5 + 3 = -2 \] Calculate: \[ (-2)^2 + (-2)(1) + 1^2 = 4 - 2 + 1 = 3 \neq 37 \] Incorrect.

Final answer:
\[ \boxed{3} \]