The known points P 2, 3 and the circle centered on Q x 4 2 y 2 2 9

Solution: (1) q(4,2),p(-2,-3).

q'(1，-1/2)

pq=√36+25=√61

Take PQ as the diameter, q'The original equation for the center of the circle is:

x-1) 2+(y+1 2) 2=61 42) is the tangent.

Proof: In APQ, AQ'=pq'= 2pq apq is a right triangle.

aq ap is the tangent of the circle q.

In the same way, pb is the tangent of the circle q.

3) AB is the circle q and the circle q'of the public chords.

Align the two circle equations.

x-4)^2+(y-2)^2=9 ①

x-1) 2+(y+1 2) 2=61 4. 6x+5y-25=0

The equation for the straight line ab is:

6x+5y-25=0

Solution: 1) Graph the connection of P and Q points, take the midpoint M of the line segment Pq as the center of the circle, and the length of the line segment MP as the radius to make a circle, that is, the circle M is found

Equation Solving Process:

p coordinates (-2, -3).

According to the equation for the circle q, the q coordinate is (4,2).

Then the coordinates of the M point are (1, and the length of the line segment PQ is [(-2-4) 2+(-3-2) ]1 2) = root number 61 circle m radius is the root number (61 4).

The equation for the circle m is (x-1) 2+(y+

The circle q and the circle m cannot be tangent.

The equation of the circle m and the circle q are combined, and the ends of (x-1) 2+(y+ equal sign are subtracted from the two ends of the equal sign of (x-4) 2+(y-2) 2=r 2 respectively, and 6x+5y=34-r 2 is the linear equation.

Because when this line intersects the circle m, when it passes through the midpoint of the circle m is the diameter, |ab|The length can be taken to the maximum, so the point m(1, substituting the equation, solving r 2=, and substituting the linear equation to obtain.

6x+5y=, i.e., 6x+5y= is the equation for the straight line.

It is known that the circle q: (x+2)2+y2=64, p(2,0), and the point m is any point on the circle + q, the line.

Hello, the circle q is (x+2)2+y2=64, then p(2,0) is not the point on this circle!

Straight line PQ: fiber head 2x+3y-2=0

Simultaneous equation with vertical force: 2x+3y-2=0

2x-y-2=0

Center of the circle: x=1 y=0 (1,0).

op^2=13

The bridge equation for a circle: (x-1) 2+y 2=13

Straight line y x Hui Jin 2 with.

Circle: x Song Da 2 y 2 2, intersect at p and q points, find 丨pq丨 ?

Substitute y x 2 into x 2 y 2 2.

x 2 (x 2) 2, 2x 2 2 2x 0, solution.

x1 a 2,x2 0,p(一 野碧vertical2,0),q(0, 2), so we get: 丨pq丨 2.

Solution: 1) Graph the connection of P and Q points, take the midpoint M of the line segment Pq as the center of the circle, and the length of the line segment MP as the radius to make a circle, that is, the circle M is found

Equation Solving Process:

p coordinates (-2, -3).

According to the equation for the circle q, the q coordinate is (4,2).

Then the coordinates of the M point are (1, and the length of the line segment PQ is [(-2-4) 2+(-3-2) ]1 2) = root number 61 circle m radius is the root number (61 4).

The equation for the circle m is (x-1) 2+(y+

The circle q and the circle m cannot be tangent.

The equations of the circle m and the circle buried q are concatenated, and the ends of the (x-1) 2+(y+equal sign are subtracted from the ends of the equal sign of (x-4) 2+(y-2) 2=r 2 respectively, and 6x+5y=34-r 2 is the linear equation.

Because when this line intersects the circle m, when it passes through the midpoint of the circle m is the diameter, |ab|The length can be taken to the maximum, so the point m(1, the crack ant is substituted into the equation, and r 2= is solved, and then the source balance of the linear equation is substituted to obtain.

6x+5y=, i.e., 6x+5y= is the equation for the straight line.

Solution: 1) Graph the connection of P and Q points, take the midpoint M of the line segment Pq as the center of the circle, and the length of the line segment MP as the radius to make a circle, that is, the circle M is found

Equation Solving Process:

p coordinates (-2, -3).

According to the equation for the circle q, the q coordinate is (4,2).

Then the coordinates of the M point are (1, and the length of the line segment PQ is [(-2-4) 2+(-3-2) ]1 2) = root number 61

The radius of the circle m is the root number (61 4).

The equation for the circle m is (x-1) 2+(y+

2) The circle q and the circle m cannot be tangent.

3) The equations of the circle m and the circle q are combined, and the ends of (x-1) 2+(y+ equal sign are subtracted from the ends of (x-4) 2+(y-2) 2=r 2 respectively, and 6x+5y=34-r 2 is the linear equation.

Because when this line intersects the circle m, when it passes through the midpoint of the circle m is the diameter, |ab|The length can be taken to the maximum, so the point m(1, substituting the equation, solving r 2=, and substituting the linear equation to obtain.

6x+5y=, i.e., 6x+5y= is the equation for the straight line.

1.The center q of the circle is (m-1,3m).

pq:y+3=[(3m+3) (m-1+2)](x+2) (m≠-1) (two-point).

i.e. y=3x+3

When m=-1, p and q coincide, and q is also on the straight line y=3x+3.

q is on the fixed line y=3x+3 over p.

2.The circle with PQ as the diameter passes through the origin, that is, the OP vertical OQ vector product -2(m-1)-3*3m=0

m=2/11

q（m-1,3m）

That is, satisfiing: x=m-1, y=3m

Solution: y=3x+3, pass the point p

Therefore, q must be on a definite line through the point p, and the equation of the straight line is: y=3x+3.

Since the center of the circle is on the line x+y=0, the coordinates of the center of the circle can be set as (a,-a) and the radius can be denoted as r

Then there is the equation for the circle m: (x-a) +y+a) = r, trace shed.

Put the dot (-2,4).(0,2) The algebraic plexus and into the equation yield:

20+12a+2a=r, 4+4a+2a=r, solution: a=2, r= 20, so the equation for the circle is.

x-2) +y+2) =20, I hope it helps, hope, thank you.

The circle collapses at the perpendicular bisector of PQ.

Above. The slope of pq is (4-2) (2-0)=-1, so the slope of the blown bisector is 1

The pq midpoint is (-1,3).

So it's x-y+4=0

The center of the circle is also x+y=0

then x=-2 and y=2

Center of the circle c(-2,2).

then r = pc = 4

The circle of the sail is (x+2) +y-2) =4