To find the length of the direct common tangent between two externally touching circles, we can use the formula for the length of a tangent between two circles with radii \( r_1 \) and \( r_2 \) and distance \( d \) between their centers:
\( L = \sqrt{d^2 - (r_1 + r_2)^2} \).
Here, the given values are:
Since the circles touch externally, the distance between their centers \( d = r_1 + r_2 = 4 + 9 = 13 \, \text{cm} \). Substituting these values into the tangent length formula gives:
\[ L = \sqrt{13^2 - (4 + 9)^2} \]
\[ L = \sqrt{169 - 169} \]
\[ L = \sqrt{0} \]
\[ L = 0 \]
Thus, the length of the direct common tangent is \( 0 \) cm.
Let $C$ be the circle $x^2 + (y - 1)^2 = 2$, $E_1$ and $E_2$ be two ellipses whose centres lie at the origin and major axes lie on the $x$-axis and $y$-axis respectively. Let the straight line $x + y = 3$ touch the curves $C$, $E_1$, and $E_2$ at $P(x_1, y_1)$, $Q(x_2, y_2)$, and $R(x_3, y_3)$ respectively. Given that $P$ is the mid-point of the line segment $QR$ and $PQ = \frac{2\sqrt{2}}{3}$, the value of $9(x_1 y_1 + x_2 y_2 + x_3 y_3)$ is equal to
The length of the latus-rectum of the ellipse, whose foci are $(2, 5)$ and $(2, -3)$ and eccentricity is $\frac{4}{5}$, is
Let the three sides of a triangle are on the lines
\(
4x - 7y + 10 = 0,\quad x + y = 5,\quad 7x + 4y = 15
\).
Then the distance of its orthocentre from the orthocentre of the triangle formed by the lines
\(
x = 0,\quad y = 0,\quad x + y = 1
\)
is