If the straight line 3x + 4y = k touches the | vedclass

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

Question

(C) Given the circle equation: $x^2 + y^2 = 16x$ Rearranging the terms: $x^2 - 16x + y^2 = 0$ Completing the square: $(x - 8)^2 + y^2 = 64$ Thus,the c

Vedclass · Accepted Answer

$-16, 64$

Vedclass · Answer

(C) Given the circle equation: $x^2 + y^2 = 16x$ Rearranging the terms: $x^2 - 16x + y^2 = 0$ Completing the square: $(x - 8)^2 + y^2 = 64$ Thus,the center is $(8, 0)$ and the radius $r = 8$. Since the line $3x + 4y - k = 0$ touches the circle,the perpendicular distance from the center $(8, 0)$ to the line must equal the radius $r = 8$. Using the distance formula: $d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}$ $8 = \frac{|3(8) + 4(0) - k|}{\sqrt{3^2 + 4^2}}$ $8 = \frac{|24 - k|}{\sqrt{25}}$ $8 = \frac{|24 - k|}{5}$ $|24 - k| = 40$ Case $1$: $24 - k = 40 \Rightarrow k = -16$ Case $2$: $24 - k = -40 \Rightarrow k = 64$ Therefore,the values of $k$ are $-16$ and $64$.

If the straight line $3x + 4y = k$ touches the circle $x^2 + y^2 = 16x$,then the value of $k$ is

Similar Questions

The equations of the tangents to the circle $x^{2}+y^{2}=13$ at the points whose abscissa is $2$ are:

Let a circle of radius $4$ and the ellipse $15x^2 + 19y^2 = 285$ be concentric. What is the angle that the common tangents make with the minor axis of the ellipse?

If the line $(x + g) \cos \theta + (y + f) \sin \theta = k$ is tangent to the circle $x^2 + y^2 + 2gx + 2fy + c = 0$,then:

The length of the tangent drawn from the mid-point of the line joining the origin and the point $(4, -4)$ to the circle $2x^2 + 2y^2 - y = 0$ is

If $O$ is the origin and $OP$,$OQ$ are tangents to the circle $x^2 + y^2 + 2gx + 2fy + c = 0$,the circumcentre of the triangle $OPQ$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module