Senior 1 math straight line and circle practice problem solving, urgent need to solve in detail

Solution: (1) Let the equation of the circle: (x-a) 2+(y-b) 2=r 2

a(2,3),b(0,3) are substituted into the equation to obtain a=1,b 2-6b+10=r 2

Since the circle is tangent to the straight line x+y-5=0, the distance from the point (1,b) to the line is the radius r, i.e., b 2-8b+16=2r 2

Coupled with the above equation, we get: b = 2, r = 2

2) Let p(c,d) meet the distance from c+d+1= point to the center of the circle d= [(c-1) 2+(d-2) 2].

pq 2=d 2-r 2 is pq 2=d 2-2, from which it can be seen that when d is the smallest, pq is also taken to the minimum value, so the intersection of the perpendicular line from the center of the circle to the line x+y+1=0 and the straight line x+y+1=0 is the p point, so the coordinates of the p point can be found: p(-1,0), so the slope of the pq straight line k=2-3 (according to the pq equation y-0=k(x+1), the distance from the center of the circle o to pq is the radius).

15(1) Let the equation of the circle be (x-a) 2+(y-b) 2=r 2, the center of the circle is (a, b), substitute the two points of a and b to get the two equations, and the distance from the center of the circle to the tangent is the radius to get the equation (a+b-5) root number 2, and form a system of equations with the above two equations to get the answer.

How can you learn this?

Because ac丄ap, the area of the quadrilateral paob = ao*pa=2pa, so pa is the smallest, and pa 2=po 2-ao 2, so po is the smallest, at this time po is a straight line l, the straight line po equation is the chain is not 2x-y=0, and the cp coordinate is (-6 5, -12 5), and the equation for the circle with p as the center and pa as the radius is.

x+6 5) 2+(y+12 5) 2=pa 2=po 2-ao 2=(-6 5) 2+(-12 5) 2-4, subtract from the circle o equation to get the ab equation is.

12x 5+24y 5=-8, simplified to 3x+6y+10=0.

When intersected, the distance from the center of the circle to the straight line is less than the radius.

So |0+0-1|Root number (A 2+B 2) < 1 root number (A 2+B 2) > 0

Multiply the root number on both sides (A 2 + B 2).

Root number (A2+B2) >1

The root number (A 2 + B 2) is the distance from P to the center of the circle, which is greater than 1, that is, greater than the radius, so P chooses B outside the park

1.(x-2)^2 + y+2)^2=82.The straight line passes through the point m(1 2,-1 2), and the chord is shortest when the slope of the chord is 1.

The slope is -m (m+1)=1 and m=-1 2

Distance from m to the center of the circle: 3 root 2

The radius of the circle is 2 times the root 2

Pythagorean theorem: Shortest length = 2 * root number (8 - 9 2) = root number 14 is right?

Solution: From the curve equation, we can get x 2 + y 2 = 1 (y 0), the image of the semicircle below the x-axis, as shown in the figure, the combination of numbers and shapes is solved, the slope of pb is k1 = 1, and the slope of kc is 4 3 (according to the distance from the center of the circle to the straight line is equal to the radius 1), so there are two different intersection points, and the value range of k is [1, 4 3).

Let the equation for the straight line be y=x+b

Substituting the equation can be obtained.

2x 2+2(b+1)x+(b 2+4b-4)=0, let a, the coordinates of b be (x1,y1)(x2,y2)x1+x2=-(b+1).

x1x2=(b^2+4b-4)/2

The circle with ab as the diameter passes the origin.

Then AO is perpendicular to BO

Vector ao=(x1,y1), vector bo=(x2,y2), x1x2+y1y2=0

x1x2+(x1+b)(x2+b)=2x1x2+b(x1+x2)+b^2=0

b^2+4b-4)-b(b+1)+b^2=0b^2+3b-4=0

b = 1 or b = -4

The equation 2x 2+2(b+1)x+(b 2+4b-4)=0 verifies that there are two solid roots.

So the equation for the sell-off line is y=x+1, y=x-4

The core of this problem is the use of vector knowledge to solve the problem.

1) Suppose the perpendicular coordinates are m(x,y), the vector am=(x-4,y), the vector bm=(x+2,y+2), the vector cm=(x-1,y-6), the vector bc=(3,8), the vector ac=(-3,6), and the vector ba=(6,2), by the perpendicular property, there are:

AM * BC = 0 BM * AC = 0 cm * BA = 0, solve the equation and get: m(20 7, 3 7).

2) Assuming that the outer center is n(x,y), because the outer center is the intersection of the perpendicular bisector of the three edges, let the midpoint of the BC edge be A, the midpoint of the AC edge be B, and the midpoint of the AB side be C. then a(-1 2,2) b(5 2,3) c(1,-1).

Vector an(x + 1 2 , y - 2) vector bn(x - 5 2, y-3) vector cn(x - 1, y + 1), so there is also:

An * bc = 0 bn * ac = 0 cn * ba = 0, solve the equation and get x = 1 14 , y = 25 14

The square of the distance from point n (1 14 , 25 14) to any point in a, b, and c is 3650 196

Therefore, the equation for the resulting circle is: (x - 1 14) 2 + y - 25 14) 2 = 3650 196

That, I calculated it in a hurry, and it's likely that the answer is not right......But the way to solve this kind of problem with vectors is the same.