The radius of the circle having its centre at $(0, 3)$ and passing through the foci of the ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$ is:

  • A
    $3$
  • B
    $3.5$
  • C
    $4$
  • D
    $\sqrt{12}$

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Answer the following by appropriately matching the lists based on the information given in the paragraph.
Let the circles $C_1: x^2+y^2=9$ and $C_2: (x-3)^2+(y-4)^2=16$ intersect at the points $X$ and $Y$. Suppose that another circle $C_3: (x-h)^2+(y-k)^2=r^2$ satisfies the following conditions:
$(i)$ The centre of $C_3$ is collinear with the centres of $C_1$ and $C_2$.
$(ii)$ $C_1$ and $C_2$ both lie inside $C_3$.
$(iii)$ $C_3$ touches $C_1$ at $M$ and $C_2$ at $N$.
Let the line through $X$ and $Y$ intersect $C_3$ at $Z$ and $W$,and let a common tangent of $C_1$ and $C_3$ be a tangent to the parabola $x^2=8 \alpha y$.
There are some expressions given in $List-I$ whose values are given in $List-II$ below:
$List-I$$List-II$
$(I) \ 2h + k$$(P) \ 6$
$(II) \ \frac{\text{Length of } ZW}{\text{Length of } XY}$$(Q) \ \sqrt{6}$
$(III) \ \frac{\text{Area of triangle } MZN}{\text{Area of triangle } ZMW}$$(R) \ \frac{5}{4}$
$(IV) \ \alpha$$(S) \ \frac{21}{5}$
$(T) \ 2\sqrt{6}$
$(U) \ \frac{10}{3}$

$(1)$ Which of the following is the only $INCORRECT$ combination?
$(1) (IV), (S) \quad (2) (IV), (U) \quad (3) (III), (R) \quad (4) (I), (P)$
$(2)$ Which of the following is the only $CORRECT$ combination?
$(1) (II), (T) \quad (2) (I), (S) \quad (3) (I), (U) \quad (4) (II), (Q)$

If two chords,each bisected by the $x$-axis,can be drawn to the circle $2(x^2 + y^2) - 2ax - by = 0$ $(a \ne 0, b \ne 0)$ from the point $P(a, b/2)$,then:

Let $r$ be the radius of the circle,which touches the $x$-axis at point $(a, 0)$,where $a < 0$,and the parabola $y^2 = 9x$ at the point $(4, 6)$. Then $r$ is equal to . . . . . . .

Let the parabola $y=x^2+px-3$ meet the coordinate axes at the points $P, Q$ and $R$. If the circle $C$ with center at $(-1,-1)$ passes through the points $P, Q$ and $R$,then the area of $\triangle PQR$ is:

Let the point $P$ be the vertex of the parabola $y = x^2 - 6x + 12$. If a line passing through the point $P$ intersects the circle $x^2 + y^2 - 2x - 4y + 3 = 0$ at the points $R$ and $S$, then the maximum value of $(PR + PS)^2$ is:

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