If a variable point (x, y) satisfies the equa | vedclass

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

Question

(A) The given equation of the circle is $x^2 + y^2 - 8x - 6y + 9 = 0$.Comparing this with the general form $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g =

Vedclass · Accepted Answer

$[ -\frac{7}{24}, \infty )$

Vedclass · Answer

(A) The given equation of the circle is $x^2 + y^2 - 8x - 6y + 9 = 0$.Comparing this with the general form $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -4$,$f = -3$,and $c = 9$.The center of the circle is $(4, 3)$ and the radius $r = \sqrt{g^2 + f^2 - c} = \sqrt{16 + 9 - 9} = \sqrt{16} = 4$.Let $m = \frac{y}{x}$,which implies $y = mx$ or $mx - y = 0$.For the line $mx - y = 0$ to intersect the circle,the perpendicular distance from the center $(4, 3)$ to the line must be less than or equal to the radius $r = 4$.Thus,$\frac{|m(4) - 3|}{\sqrt{m^2 + (-1)^2}} \le 4$.$|4m - 3| \le 4\sqrt{m^2 + 1}$.Squaring both sides,$(4m - 3)^2 \le 16(m^2 + 1)$.$16m^2 - 24m + 9 \le 16m^2 + 16$.$-24m \le 7$,which gives $m \ge -\frac{7}{24}$.Since the line $y = mx$ passes through the origin and the circle lies entirely in the first quadrant (as the center is $(4, 3)$ and radius is $4$,$x$ ranges from $0$ to $8$ and $y$ from $-1$ to $7$),the slope $m$ can take values from $-\frac{7}{24}$ to $\infty$.

If a variable point $(x, y)$ satisfies the equation $x^2 + y^2 - 8x - 6y + 9 = 0$,then the range of $\frac{y}{x}$ is

Similar Questions

If the image of the point $(-4, 5)$ in the line $x + 2y = 2$ lies on the circle $(x + 4)^2 + (y - 3)^2 = r^2$,then $r$ is equal to:

If any tangent drawn to the ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$ touches one of the circles $x^2 + y^2 = \alpha^2$,then the range of $\alpha$ is

The straight line touching the circle $x^2+y^2-2x-3=0$ and remaining normal to the circle $x^2+y^2-4y-6=0$ is

$ax - y + c = 0$ is the equation of the common tangent to the parabola $y^2 = 8\sqrt{5}x$ and the circle $x^2 + y^2 = 1$. If this tangent makes an acute angle with the positive $X$-axis,then $a^2c^2 =$

Let $\theta$ be the acute angle between the tangents to the ellipse $\frac{x^{2}}{9}+\frac{y^{2}}{1}=1$ and the circle $x^{2}+y^{2}=3$ at their point of intersection in the first quadrant. Then $\tan \theta$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module