The line 3x + y - 5 = 0 touches a circle S at | vedclass

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(B) Given $h^2 + hk + k^2 = 37$ ... $(i)$ Radius $r = \sqrt{10}$. The equation of the circle $S$ is $(x - h)^2 + (y - k)^2 = 10$ ... $(ii)$ Since $(1,

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$3$

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(B) Given $h^2 + hk + k^2 = 37$ ... $(i)$ Radius $r = \sqrt{10}$. The equation of the circle $S$ is $(x - h)^2 + (y - k)^2 = 10$ ... $(ii)$ Since $(1, 2)$ lies on the circle,$(1 - h)^2 + (2 - k)^2 = 10$. Expanding this,$1 - 2h + h^2 + 4 - 4k + k^2 = 10$,which simplifies to $h^2 + k^2 - 2h - 4k = 5$. Substituting $h^2 + k^2 = 37 - hk$ from $(i)$ into this equation: $37 - hk - 2h - 4k = 5 \Rightarrow hk + 2h + 4k = 32$ ... $(iii)$ The perpendicular distance from the centre $(h, k)$ to the tangent line $3x + y - 5 = 0$ is equal to the radius $\sqrt{10}$: $\frac{|3h + k - 5|}{\sqrt{3^2 + 1^2}} = \sqrt{10} \Rightarrow |3h + k - 5| = 10$. Assuming $3h + k - 5 = 10$,we get $3h + k = 15 \Rightarrow k = 15 - 3h$. Substituting $k = 15 - 3h$ into $(iii)$: $h(15 - 3h) + 2h + 4(15 - 3h) = 32$ $15h - 3h^2 + 2h + 60 - 12h = 32$ $-3h^2 + 5h + 28 = 0 \Rightarrow 3h^2 - 5h - 28 = 0$. Solving the quadratic equation $(3h + 7)(h - 4) = 0$,we get $h = 4$ or $h = -7/3$. If $h = 4$,then $k = 15 - 3(4) = 3$. Thus,$k = 3$.

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

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