The equation for a straight line tangent to the crossing point 2 3 and the circle X 1 square y squar

Hello! Let the equation of the straight line be y=kx+b, i.e., kx-y+b=0, the center of the circle is (1,0), and the radius is 1

d=|ax+by+c|/√（a²+b²）=|k+b|/√(k²+1)=1...k²+1=（k+b）²

Straight line through point p(2,3).

2k+b=3...

Two-way (or directly substituting relational expressions).

b=1/3 k=4/3

The equation for a straight line is y=4 3x+1 3

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Solution: Let the equation for the straight line of the point (2,3) be y-3 =k(x-2).

y=k(x-2)+3

Bring in the meta equation (x-1) 2+y 2 =1.

x-1)^2+[kx+(3-2k)]^2 =1

x^2-2x+1 +k^2 x^2 +2k(3-2k)x+(3-2k)^2 -1 =0

k^2+1)x^2 -2(2k^2 -3k+1) x +(3-2k-1)(3-2k+1)=0

k^2+1)x^2 -2(2k^2 -3k+1) x +4(k^2 -3k +2)=0

It is obtained by factoring by cross multiplication.

k^2+1)x - 2(k^2-3k+2)](x-2) =0

Because of the tangent, there can only be one solution of x=2, that is, the solution of (k 2+1)x - 2(k 2-3k+2)=0 should also be x=2

k^2+1)2- 2(k^2-3k+2)=0

k^2+1= k^2-3k+2

k=1 3, so the tangent equation is y=1 3(x-2)+3 =x 3 +7 3

If you learn the derivative, the problem is very simple, you only need to find the derivative of the other equation at the point (2, 3), and you can directly get the slope of 1 3

It can be seen that the coordinates of the center of the circle are (1,0) and the radius is 1, and the analytical slope of the line where the center of the circle and the point (2,3) are located is set to a, then the slope of the straight line is set to b, then a is multiplied by b=-1, and there is a slope, and the point (2,3) is crossed

You can find the analytic formula for the straight line--

Because a(2,1) and x 2+y 2=1 are judged to be tangential, so when the slope is zero, the equation is y=1, when the oblique brother rate is not zero and exists, let the slope be k, so the equation is y-1=k(x-2), substituting x 2+y 2=1, the solution is k= 3, in summary, the equation is y=1 or y= 3x-2 3+1 is dead n multi-brain fine digging cell o( o, I hope it helps you.

Let the equation for the straight line be .

y+1=k(x-2)

Namely. kx-y-2k-1=0

Using the formula for the distance from the point to the line, the distance from the center of the circle (1,1) to the line is equal to the radius.

Namely. k-1-2k-1|k 2 + 1) = 5 to solve k.

If the center of the circle is on the straight line y=-2x, then the center of the circle can be set to be (m,-2m) then the standard equation of the circle can be expressed as: (x-m) 2+(y+2m) 2=r 2 and the circle is tangent to the straight line y=1-x, then the distance from the straight line d=|m-2m-1|/√2=r

i.e.: (m+1) 2 2=r 2

And because the tangent point of the stool is on the circle, (2,-1) satisfies the equation of the circle, and substituting it gets:

So, m=1

Then the standard equation for a circle is:

x-1)^2+(y+2)^2=2

If you don't understand, please ask, satisfied, thank you

Let the straight line be y=k(x-3)+2

The distance from the center of the circle to the straight line is equal to the radius.

So |k(0-3)+2-0|k 2 + 1) = 1 solution k = (-3 3) 4

So the equation for the straight line is y=(-3 3)(x-3) 4+2(2) respectively with the straight line.

y=-2xand.

y=x 2 tangent.

Let the center of the circle be (a,b).

then |2a+b|/√4+1)=|a-2b|1+4) so a=-3b or b=3a

When a=-3b.

Substituting the above equation gives a radius of |2a+b|/√4+1)=(5)*|b|Circle over the dot (3, 2).

Thus (a-3) 2+(b-2) 2=5b 2 simultaneous a=-3b finds no solution.

When b = 3a.

Substituting the above equation gives a radius of |2a+b|/√4+1)=(5)*|a|Circle over the dot (3, 2).

So (a-3) 2+(b-2) 2=5a 2 joins b=3a to get 5a 2-18a+13=0

The solution yields a= or a=1

a=b=3a=, the radius is (5)*|a|= So the equation for the circle is (

When a=1, b=3a=3, the radius is (5)*|a|= 5, so the equation for a circle is (x-1) 2+(y-3) 2=5

Drawing, first roughly draw the required straight line and circle, and then connect the center of the circle and the tangent point, which is to form a right-angled triangle, the radius is one, then a right-angled side is one, the hypotenuse is 2, the angle between the straight line and the x-axis is 180 degrees - 30 degrees = 150 degrees, then the slope is tan30 degrees = one-third of the root number, and then according to y=kx+b, substitute the point (2, 0), you can get b, and you can find it.

Let the slope of the straight line be k, then the equation of the straight line is y=k(x-2) the straight line is tangent to the circle x square + y square = 1, and the distance from the origin of the circle to the straight line is the radius 1 ,|2k|/√(k^2+1)=1

4k^2=k^2+1

k^2=1/3

k = 3 3 or - 3 3

y=3 3 (x-2) or.

y=-√3/3(x-2)

Let the equation for this line be: y

k(x2), i.e.: kx

y-2k=0

So, 丨-2k丨 (k +1)1 is solved, k

3 The equation for this straight line is: y

Substituting the point coordinates into the equation of the circle, we can see that the point is on the circle, that is, there is only one tangent line, and the slope of the straight line passing through the point (2,-1) and the center of the circle is (-1-1) (2-1)=-2, then the slope of the straight line tangent to the circle through the point (2,-1) is 1 2, substituting the point (2,-1) into y+1=(1 2)(x-2), and the tangent straight line equation is y=x 2-2

This point is on a circle and the slope is (1-(-1)) (1-2)= -2, then the slope of the straight line is 1 2, then the equation for the straight line is y=x 2-2

Let the slope of the straight line be k

Then the equation for the straight line is y-4=k(x+3).

kx-y+3k+4=0

The coordinates of the center of the circle are (1,1) and the radius is 5

To be tangent, the distance from the center of the circle to the straight line is equal to the radius.

So |k-1+3k+4|(k +1) = 5 to get k = -2 or -2 11

So the equation for a straight line is y=-2x-2 or y=-2x, 11+38, 11, i.e., 2x+y+2=0 or 2x+11y-38=0

Let the slope be k, the distance from the center of the circle (1,1) to the straight line y-4=k(x+3) = 5, find the equation of the straight line k and the equation of the circle to be concatenated, delta = 0, solve k

k = 4/3 can't solve -2 and that.

(1) Let the straight line be y=k(x-3)+2

The distance from the center of the circle to the straight line is equal to the radius.

So |k(0-3)+2-0|k 2 + 1) = 1 solution k = (-3 3) 4

Therefore, the equation of the straight line is y=(-3 3)(x-3) 4+2(2), tangent to the straight lines y=-2x and y=x 2, respectively, and the center of the circle is (a,b).

then |2a+b|/√(4+1)=|a-2b|1+4) so a=-3b or b=3a

When a=-3b.

Substituting the above equation gives a radius of |2a+b|/√(4+1)=(√5)*|b|Circle over the dot (3, 2).

Thus (a-3) 2+(b-2) 2=5b 2 simultaneous a=-3b finds no solution.

When b = 3a.

Substituting the above equation gives a radius of |2a+b|/√(4+1)=(√5)*|a|Circle over the dot (3, 2).

So (a-3) 2+(b-2) 2=5a 2 joins b=3a to get 5a 2-18a+13=0

The solution yields a= or a=1

a=b=3a=, the radius is (5)*|a|= So the equation for the circle is (

When a=1, b=3a=3, the radius is (5)*|a|= 5, so the equation for a circle is (x-1) 2+(y-3) 2=5