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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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,

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(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 $k =$

Similar Questions

The angle between the tangents drawn from the origin to the circle $(x - 7)^2 + (y + 1)^2 = 25$ is:

The angle between the tangents drawn from the origin to the circle $x^2+y^2+4x-6y+4=0$ is

The equations of two diameters of a circle are $2x - 3y = 5$ and $3x - 4y = 7$. The line joining the points $\left(-\frac{22}{7}, -4\right)$ and $\left(-\frac{1}{7}, 3\right)$ intersects the circle at only one point $P(\alpha, \beta)$. Then $17\beta - \alpha$ is equal to

The focal chord of the parabola $(y - 2)^2 = 16(x - 1)$ is a tangent to the circle $x^2 + y^2 - 14x - 4y + 51 = 0$. Then the slope of the focal chord can be:

The tangent to the parabola $y = x^2 + 6$ at the point $(1, 7)$ touches the circle $x^2 + y^2 + 16x + 12y + c = 0$ at which point?

Vedclass Test Series

Exam Paper Generator

Online Exam Module