If the power of a point (4, 2) with respect t | vedclass

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

Question

(A) The equation of the circle is $x^2 + y^2 - 2\alpha x + 6y + \alpha^2 - 16 = 0$. Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -\alpha

Vedclass · Accepted Answer

$16 + 4\sqrt{6}$

Vedclass · Answer

(A) The equation of the circle is $x^2 + y^2 - 2\alpha x + 6y + \alpha^2 - 16 = 0$. Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -\alpha, f = 3, c = \alpha^2 - 16$. The power of a point $(x_1, y_1)$ with respect to a circle $S = 0$ is $S(x_1, y_1) = x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c$. Given power is $9$ at $(4, 2)$: $4^2 + 2^2 - 2\alpha(4) + 6(2) + \alpha^2 - 16 = 9$ $16 + 4 - 8\alpha + 12 + \alpha^2 - 16 = 9$ $\alpha^2 - 8\alpha + 7 = 0$ $\Rightarrow (\alpha - 1)(\alpha - 7) = 0$ $\Rightarrow \alpha = 1, 7$. Case $1$: $\alpha = 1$,circle is $x^2 + y^2 - 2x + 6y - 15 = 0$. $x$-intercept: set $y = 0$ $\Rightarrow x^2 - 2x - 15 = 0$ $\Rightarrow (x-5)(x+3) = 0$ $\Rightarrow x = 5, -3$. Length $= |5 - (-3)| = 8$. $y$-intercept: set $x = 0$ $\Rightarrow y^2 + 6y - 15 = 0$ $\Rightarrow y = -3 \pm \sqrt{9 + 15} = -3 \pm 2\sqrt{6}$. Length $= |(-3 + 2\sqrt{6}) - (-3 - 2\sqrt{6})| = 4\sqrt{6}$. Case $2$: $\alpha = 7$,circle is $x^2 + y^2 - 14x + 6y + 33 = 0$. $x$-intercept: set $y = 0$ $\Rightarrow x^2 - 14x + 33 = 0$ $\Rightarrow (x-11)(x-3) = 0$ $\Rightarrow x = 11, 3$. Length $= |11 - 3| = 8$. $y$-intercept: set $x = 0 \Rightarrow y^2 + 6y + 33 = 0$. Discriminant $D = 36 - 4(33) Total sum of lengths $= 8 + 4\sqrt{6} + 8 = 16 + 4\sqrt{6}$.

If the power of a point $(4, 2)$ with respect to the circle $x^2 + y^2 - 2\alpha x + 6y + \alpha^2 - 16 = 0$ is $9$,then the sum of the lengths of all possible intercepts made by such circles on the coordinate axes is:

Similar Questions

$A$ semi-circle of diameter $1$ unit sits at the top of a semi-circle of diameter $2$ units. The shaded region inside the smaller semi-circle but outside the larger semi-circle is called a lune. The area of the lune is

If a straight line through $C(-\sqrt{8}, \sqrt{8})$ making an angle of $135^\circ$ with the $x$-axis cuts the circle $x = 5\cos\theta, y = 5\sin\theta$ at points $A$ and $B$,then the length of $AB$ is

Let $6$ and $8$ be the $X$ and $Y$-intercepts made by the circle $S \equiv x^2+y^2+2gx+2fy+c=0$ respectively. If $gx+fy+1=0$ is a line passing through the point $(1, -1)$,then the radius of the circle $S=0$ is

Let the equation of the circle,which touches the $x$-axis at the point $(a, 0), a > 0$ and cuts off an intercept of length $b$ on the $y$-axis,be $x^2 + y^2 - \alpha x + \beta y + \gamma = 0$. If the circle lies below the $x$-axis,then the ordered pair $(2a, b^2)$ is equal to:

From three non-collinear points,we can draw:

Vedclass Test Series

Exam Paper Generator

Online Exam Module