Answer:
Note that the tangents to the circles at A and B intersect at a point Z on XY by radical center. Then, since∠ZAB=∠ZQA and ∠ZBA=∠ZQB. ∠AZB+∠AQB=∠AZB+∠ZAB+∠ZBA=180°. ∴ZAQB is cyclic.But if O is the center of w, clearly ZAOB is cyclic with diameter ZO, so ∠ZQO is 90° ⇒Q is the mid-point of XY.Then, by Power of a Point, PY· PX = PA · PB = 15 and it is given that PY+PX = 11. Thus,PX=(11±)/2. So, PQ=
, PQ²=
. Thus, the answer is 61+4=65.
Step-by-step explanation:
Answer:
y = x - 3
Step-by-step explanation:
The equation of a line in slope- intercept form is
y = mx + c ( m is the slope and c the y- intercept )
To calculate m use the slope formula
m = ( y₂ - y₁ ) / ( x₂ - x₁ )
with (x₁, y₁ ) = (- 22, - 14) and (x₂, y₂ ) = (- 18, - 12)
m = =
=
y = x + c ← is the partial equation
To find c substitute either of the 2 points into the partial equation
using (- 22, - 14), then
- 14 = - 11 + c ⇒ c = - 14 + 11 = - 3
y = x - 3 ← in slope- intercept form