A circle S touches the Y-axis at (0,3) and ma | vedclass

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

Question

(C) The equation of a circle that touches the $Y$-axis at $(0,3)$ is given by $(x-a)^2 + (y-3)^2 = a^2$,which simplifies to $x^2 + y^2 - 2ax - 6y + 9

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(C) The equation of a circle that touches the $Y$-axis at $(0,3)$ is given by $(x-a)^2 + (y-3)^2 = a^2$,which simplifies to $x^2 + y^2 - 2ax - 6y + 9 = 0$.Since the length of the intercept on the $X$-axis is $8$ units,we use the formula $2\sqrt{g^2 - c} = 8$,where $g = -a$ and $c = 9$.$2\sqrt{(-a)^2 - 9} = 8$ $\Rightarrow \sqrt{a^2 - 9} = 4$ $\Rightarrow a^2 - 9 = 16$ $\Rightarrow a^2 = 25$ $\Rightarrow a = \pm 5$.Since the centre $C(a, 3)$ lies in the second quadrant,$a$ must be negative,so $a = -5$.Thus,the centre is $C(-5, 3)$.The distance of $C(-5, 3)$ from the point $(-2, -1)$ is $\sqrt{(-2 - (-5))^2 + (-1 - 3)^2} = \sqrt{(3)^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.

$A$ circle $S$ touches the $Y$-axis at $(0,3)$ and makes an intercept of length $8$ units on the $X$-axis. If the centre $C$ of the circle $S$ lies in the second quadrant,then the distance of $C$ from the point $(-2,-1)$ is

Similar Questions

The length of the chord of the circle $x^{2}+y^{2}+3x+2y-8=0$ intercepted by the $y$-axis is

The acute angle between the curves $x^2+y^2=x+y$ and $x^2+y^2=2y$ is

If $5x - 12y + 10 = 0$ and $12y - 5x + 16 = 0$ are two tangents to a circle,then the radius of the circle is

Let $B$ be the centre of the circle $x^{2}+y^{2}-2x+4y+1=0$. Let the tangents at two points $P$ and $Q$ on the circle intersect at the point $A(3,1)$. Then $8 \left(\frac{\text{Area } \triangle APQ}{\text{Area } \triangle BPQ}\right)$ is equal to:

If ${c^2} > {a^2}(1 + {m^2})$,then the line $y = mx + c$ will intersect the circle ${x^2} + {y^2} = {a^2}$ at:

Vedclass Test Series

Exam Paper Generator

Online Exam Module