Circle
A circle is the locus of all points that are a fixed distance (radius) from a fixed point (center).
Standard form:
$$ (x - h)^2 + (y - k)^2 = r^2 $$
(h,k) is the center of the circle
A circle is the locus of all points that are a fixed distance (radius) from a fixed point (center).
Standard form:
(h,k) is the center of the circle
Find the equation of the circle with center at (-3,8) and tangent to the line x-y+5=0.
Determine the length of the tangent to the circle x2+y2-4x-5=0 from (8,-2).
Find the locus of a moving point which forms a triangle of area 21 square units with the point (2, -7) and (-4, 3).
Ans. 5x+3y+32=0 or 5x+3y-10=0
There is a fixed circle having a radius of 6 with center at (10, 12). Find the equation of the curve connecting the centers of all circles that are tangent to the fixed circle and the x axis.
Observation: The equation is quadratic in $x$ but only first degree in $y$. A curve of the form $$Ax^2 + Ey + F = 0$$ is a parabola. Hence, the locus is a parabola.
Given the equations of two circles:
C₁: x² + y² + 2x + 4y − 3 = 0
C₂: x² + y² − 8x − 6y + 7 = 0
Determine the equation of the common tangent at their point of contact.
Step 1: Subtract the equations of the two circles
C₁: x² + y² + 2x + 4y − 3 = 0
C₂: x² + y² − 8x − 6y + 7 = 0
--------------------------------
10x + 10y − 10 = 0
Simplifying:
$\boxed{x + y = 1}$This line is the common tangent at the point of contact.
When two circles are tangent internally or externally, their point of tangency lies on a line that is perpendicular to the line joining the centers. The key idea: The difference of the equations of two circles gives the radical axis—the locus of points having equal power with respect to both circles.
If two circles intersect at two points, the radical axis is their common chord. But if the circles touch at exactly one point (tangent), the radical axis collapses to the common tangent at the point of contact.
Therefore, since subtracting C₂ from C₁ yields a linear equation, it must represent the common tangent.
A circle has its center at (2, 3). If the circle is tangent to the y-axis and a tangent line has its point of tangency at P(3, y), determine the value of y.
Step 1: Determine the radius using the tangency to the y-axis
The distance from the center $(2,3)$ to the y-axis $(x = 0)$ is: $$ r = |2| = 2 $$
Step 2: Use the right triangle formed by the center and point of tangency
Let the point of tangency be $P(3, y)$. The horizontal distance from the center to $P$ is: $$ 3 - 2 = 1 $$
Let the vertical distance be $h$. Using the Pythagorean relationship: $$ h^2 + 1^2 = r^2 $$ $$ h^2 + 1 = 4 $$ $$ h^2 = 3 $$ $$ h = \sqrt{3} $$
Step 3: Compute the y-coordinate
$$ y = 3 + h $$ $$ y = 3 + \sqrt{3} $$
Final Answer:
$$ \boxed{y = 3 + \sqrt{3}} $$
For what value of k does the circle $ (x + 2k)^2 + (y - 3k)^2 = 10 $ pass through the point (1,0)?
Substitute the point (1,0) into the circle equation:
\[ (1 + 2k)^2 + (0 - 3k)^2 = 10 \]Expand each term:
\[ (1 + 2k)^2 = 1 + 4k + 4k^2 \] \[ (-3k)^2 = 9k^2 \]Sum them:
\[ 1 + 4k + 4k^2 + 9k^2 = 10 \]Combine like terms:
\[ 13k^2 + 4k + 1 = 10 \]Move 10 to the left:
\[ 13k^2 + 4k - 9 = 0 \]Solve the quadratic:
\[ k = \frac{-4 \pm \sqrt{4^2 - 4(13)(-9)}}{2(13)} \] \[ k = \frac{-4 \pm \sqrt{16 + 468}}{26} \] \[ k = \frac{-4 \pm \sqrt{484}}{26} \] \[ k = \frac{-4 \pm 22}{26} \]Thus, the two possible values are:
\[ k = \frac{18}{26} = \frac{9}{13}, \qquad k = \frac{-26}{26} = -1 \]Final Answer:
\[ \boxed{k = -1 \quad \text{or} \quad k = \frac{9}{13}} \]
Determine the farthest distance from the point $(8,10)$ to the circle $x^2 + y^2 = 16y.$
Step 1: Rewrite the circle in standard form
$$x^2 + y^2 - 16y = 0$$ Complete the square: $$x^2 + (y^2 - 16y + 64) = 64$$ $$(y - 8)^2 + x^2 = 8^2$$ The center is $(0,8)$ and the radius is $$r = 8.$$Step 2: Find the distance from the point to the center
$$d_1 = \sqrt{(8 - 0)^2 + (10 - 8)^2}$$ $$d_1 = \sqrt{64 + 4} = \sqrt{68} = 8.25$$Step 3: Farthest distance = center-to-point distance + radius
$$d_2 = d_1 + r$$ $$d_2 = 8.25 + 8 = 16.25$$Final Answer:
$$\boxed{16.25}$$