Find the distance from the center of circle x + y square = 9 to the line 3x + 4y-1 = 0

Find the distance from the center of circle x + y square = 9 to the line 3x + 4y-1 = 0

Center (0,0)
The distance to the line 3x + 4y-1 = 0 = | 0 + 0-1 | / radical (3 ^ 2 + 4 ^ 2) = 1 / 5

Given the circle C: (x-a) ^ 2 + (Y-A) ^ 2 = a ^ 2 and the straight line L: 3x + 4Y + 3 = 0, if there are and only two points on the circle C whose distance from L is equal to 1, find the value range of A

The center of circle C is C (a, a), and the radius r = | a |
Let the distance between the line 3x + 4Y + C = 0 and the line 3x + 4Y + 3 = 0 = 1,
Then | C-3 | / 5 = 1, then C = 8 or - 2,
Therefore, according to the known conditions, the circle C intersects the straight line 3x + 4Y + 8 = 0 and is separated from 3x + 4Y-2 = 0,
Or circle C intersects the line 3x + 4Y-2 = 0 and is separated from 3x + 4Y + 8 = 0
So from the distance formula of the point to the straight line
|3a+4a+8|/5|a| ； （1）
Or 1243a + 4A + 8 / 5 > | a | and a + 4a-2 / 5

If the circle C is tangent to the line 3x-4y-18 = 0, the shortest distance between the point on circle C and the line x-4y-3 = 0 is equal to 1 x-y-3 = 0 If the straight line x-y-3 = 0 intersects with circle C, the chord length is equal to the root sign 2. Find the standard equation of circle C

It seems that I haven't learned it yet

In the plane rectangular coordinate system, the straight line y = 3x + 1______ Translation______ The linear y = 3x-4 is obtained

In the plane rectangular coordinate system, the straight line y = 3x + 1 becomes the straight line y = 3x-4, which needs to be translated down 5 units

As shown in the figure, in the plane rectangular coordinate system, the coordinates of point a are (1,0). Take a point P on the straight line y = √ 3 / 3x, so that △ OPA is an isosceles triangle. Find all the P coordinates that meet the conditions

In this paper, if the OAA | | op | (OP

Then the distance between the line a and the straight line x, Y / 4, and the straight axis A and B are satisfied respectively How many lines L satisfy the condition?

Is it four? The straight line y = - 4 / 3x + 4 intersects with X axis and Y axis at (1 / 3, O) (0,4) two points respectively. The two circles are separated and have four common tangent lines, so they are four

In the plane rectangular coordinate system, the straight line y = - 3x + 2 and the straight line y = 3x + 2 intersect at point P, and the two lines intersect with the X axis at points a and B respectively, and the origin is o (1) The coordinates of the intersection point P are obtained; (2) Judge whether △ APB is isosceles triangle and explain the reason

(1) By
y＝−3x+2
Y = 3x + 2
x＝0
Y = 2 (4 points)
So the coordinates of point P are (0, 2) (5 points)
(Note: if the drawing is done correctly through the line of point tracing in the list, the score is similar to that in Figure 4)
(2) Δ APB is isosceles triangle, reason: (6 points)
Let y = 0
-3x+2=0
X = 2
Three
So the coordinate of point a is (2)
3, 0) (7 points)
3x+2=0
X = 2
Three
So the coordinates of point B are (− 2)
3, 0) (8 points)
∴OA=0B=2
3 (10 points)
Also op ⊥ ab
/ / PA = Pb. (12 points)

As shown in the figure, in the plane rectangular coordinate system, O is the coordinate origin, the straight line y = 3x + 9 intersects with X axis and Y axis at a and C respectively, and the parabola y is shown in the figure, which is straight in the plane

Is this the question
As shown in the figure, in the plane rectangular coordinate system, O is the coordinate origin. The straight line y = 3x + 9 intersects with X axis and Y axis at two points a and C respectively. The parabola y = - 1 / 4x 2 + BX + C passes through two points a and C, and the other intersection point of X axis is point B. the moving point P starts from point a and moves along AB at the speed of 3 unit length per second to point B, and the moving point Q starts from point B and moves along BC at the speed of 3 unit length per second to point C, Starting from point C, moving point n moves towards point a at the speed of 3 √ 10 / 5 unit length per second along ca. points P, Q and N start and stop at the same time, and the movement time is t (0 ＜ t ＜ 5) seconds
(1) Find the analytic formula of parabola;
(2) Judge the large shape of △ ABC;
(3) When ⊙ o ⊙ and BC with OC as diameter intersect at point m, find out what value t is, PM is tangent to ⊙ o ′? Please explain the reason;
(4) In the process of point P, Q and N moving, is there any case where △ NCQ is a right triangle? If so, find out the corresponding t value; if not, please explain the reason
(1) At y = - 3 4
X+9
If x = 0, y = 9; if y = 0, x = 12
∴C（0,9）,B（12,0）．
Then the parabola passes through B and C, C = 9
-36 + 12b + C = 0, the solution is b = 9 4 C = 9 ﹤ y = - 1 4 x 2 + 9 4 x + 9
So let y = 0, we get - 1 4 x 2 + 9 4
x+9=0,
X 1 = - 3, x 2 = 12. A (- 3,0)
(2) When t = 3 seconds, PM is tangent to ⊙ o '
 OC is the diameter of ⊙ o ′,  OMC = 90 °.  OMB = 90 °
∵ o ′ o is the radius of ⊙ o ⊥ o ⊥ OP, ᙽ OP is the tangent line of ⊙ o ⊥ o ⊙ o ⊥ o ⊙ o ⊥ o ⊙ o ⊥ Op
And PM is the tangent of ⊙ o ′,  PM = Po.  POM = ∠ PMO
And ? POM + ∠ OBM = 90 °, PMO + ∠ PMB = 90 °, PMB = ∠ OBM. ? PM = Pb
∴PO=PB=1
At this time, t = 3 (seconds)
When t = 3 seconds, PM is tangent to ⊙ o '
(3) (1) make QD ⊥ ob at point d through point Q
∵OC⊥OB,∴QD∥OC．∴△BQD∽△BCO．∴QD OC =BQ BC ．
And ∵ OC = 9, BQ = 3T, BC = 15, ∵ QD 9 = 3T 15
It is found that QD = 9.5 t
﹤ s △ bpq = 1 2 BP · QD = - 27 10 T2 + 27 2 T, i.e. s = - 27 10 T2 + 27 2 t
S=-27
Therefore, when t = 52, s is the maximum, and the maximum value is 1358
.
② There exists the case that △ NCQ is a right triangle
∵ BC = Ba = 15,  BCA = ∠ BAC, i.e.  NCM = ∠ Cao
If △ NCQ wants to be a right triangle, ∠ NCQ ≠ 90 ° only exists ∠ NQC = 90 ° and ∠ QNC = 90 °
When ∠ NQC = 90 °, NQC = ∠ COA = 90 °, NCQ = ∠ Cao,
∴△NCQ∽△CAO．∴NC
CA = CQ Ao. ν 3 105 t 32 + 92 = 15-3t 3, t = 256
.
When ∠ QNC = 90 °, QNC = ∠ COA = 90 °, qcn = ∠ Cao,
∴△QCN∽△CAO．∴CQ AC =NC OA ．∴15-3t
32 + 92 = 3 105 T 3, t = 53
To sum up, there exists the case that △ NCQ is a right triangle, and the values of T are 256 and 53
Copied from other places, some mathematical symbols can not be typed tat

As shown in the figure, in the plane rectangular coordinate system, the straight line l1:y = 4 3x and line L2: y = KX + B intersect at point a, abscissa of point a is 3, line L2 intersects Y axis at point B, and | OA | = 1 2|OB|． (1) Try to find the function expression of line L2; (2) If the line L1 is shifted to the left by 3 units along the X axis, the intersection Y axis is at point C, and the intersection line L2 is at point D. try to find the area of △ BCD

(1) According to the meaning of the title, the abscissa of point a is 3, substituting it into the line L1: y = 43x, the ordinate of point a is 4, that is, point a (3, 4); that is, OA = 5, and | OA | = 12 | ob |. That is, OB = 10, and point B is located on the Y axis, that is, B (0, - 10); substituting the coordinates of point a and B into the line L2, we get 4 = 3K + B; - 10 = B

It is known that in the plane rectangular coordinate system, the straight line y = - 3x-3 intersects X axis at point a, intersects Y axis at point C, coordinates of point B are (3.0), parabola passes through a, B, C three It is known whether there is a point P on the parabola of point d (4. - 1), such that PD is a circle of diameter o 'passing through the origin O. if point P exists, the coordinates of the point P satisfying the conditions are obtained

B (3,0) C (0, - 3) a (- 1,0) y = ax square + BX + CC = - 30 = 9A + 3b-3 = a-b-3a-b = 33A + B = 1A = 1 b = - 2, that is, y = x-2x-3 such that PD is a circle of diameter o 'passing through the origin o, i.e., the triangle DPO is a right triangle Po + od = DP x-x + (x-2x-3) x + 16 + 1 = (x-4) + (x-2x-3-1) x + (x-4) + (X-2) x-3 + (x-4) X-2