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The linear equation passing through point m (2,1) and tangent to circle x2 + y2-6x-8y + 24 = 0 is______ ．
Circle x2 + y2-6x-8y + 24 = 0 is transformed into the standard equation as (x-3) 2 + (y-4) 2 = 1, center of circle (3, 4), radius r = 1. When the slope does not exist, x = 2 is the tangent of the circle, which satisfies the problem; when the slope exists, let the equation be Y-1 = K (X-2), that is, kx-y + 1-2k = 0 | from the center of circle to the straight line d = R, we can get | K − 3 | K2 + 1 = 1 | k = 43, | linear equation is 4x-3y-5 = 0. In conclusion, the tangent equation is x = 2 or 4x-3y-5= So the answer is: x = 2 or 4x-3y-5 = 0
Find the equation of the circle which is concentric with the circle X & sup2; + Y & sup2; - 4x + 6y3 + = 0 and passes through the point (- 1,1)
The circle X & sup2; + Y & sup2; - 4x + 6y + 3 = 0 is transformed into (X-2) & sup2; + (y + 3) & sup2; = 10. Then the original equation concentric with it is set as (X-2) & sup2; + (y + 3) & sup2; = R & sup2; the point (- 1,1) is brought into the equation (X-2) & sup2; + (y + 3) & sup2; = R & sup2; and R = 5. So the result is (X-2) & sup2; + (y + 3) & sup2; = 25
Let x ^ 2 + y ^ 2 + 6x-8y = 0, then the radius of the circle is?
(x+3)² + (y-4)²=5²
r=5
RELATED INFORMATIONS
- 1. Given that the equation of a circle is x2 + y2-6x-8y = 0, let the longest chord and the shortest chord of the circle passing through the point (3, 5) be AC and BD respectively, then the area of the quadrilateral ABCD is () A. 106B. 206C. 306D. 406
- 2. The equation of circle C is known as x ^ 2 + y ^ 2-6c-8y + 21 = 0 The equation of straight line is kx-y - (4k-3) = 0 (1) It is proved that no matter why K is a real number, the line L and C must intersect (2) Let l be at the intersection of circle C, AB, what is k, and what is the minimum value of AB
- 3. If the equation x2 + Y2 + 4mx-2y + 4m2 + 4m = 0 represents a circle, then the value range of real number m is
- 4. Find the real number m so that the straight line X-my + 3 = 0 and the circle x 2 + y 2 = 6x-5 satisfy the following conditions: (1) intersection; (2) tangency; (3) separation
- 5. If the line X-my + 3 = 0 and the circle (x-3) ^ 2 + y ^ 2 = 4 are separated, then the value range of real number m is
- 6. If the line X-my + 1 = 0 is tangent to the circle x ^ 2 + y ^ 2-2x = 0, what is the value of M
- 7. If the real number x, y satisfies X-Y + 1 > = 0, y + 1 > = 0, x + y + 1
- 8. Given that real numbers x and y satisfy 2x + y ≥ 43x + y ≤ 62x − y ≥ - 4, then the value range of Y − 2x − 2 is obtained______ ．
- 9. Given that the real number x y satisfies the constraint condition x > = 1, Y > = 0, X-Y > = 0, then the value range of Z = (Y-1) / X is? A [- 1,0] B (negative infinity, 0] C [- 1, positive infinity) d [- 1,1)
- 10. If | X-1 | + | X-2 | + | x-3 | + +|If x-2007 | is greater than or equal to a for all real numbers x, then the value range of real number a is? Well
- 11. The common tangent of circle x ^ 2 + y ^ 2 = 1 and circle x ^ 2 + y ^ 2-6x-8y + 9 = 0 are
- 12. The equation x & sup2; + Y & sup2; + ax + 2ay + 2A & sup2; + A-1 = 0 indicates the value range of garden rule a
- 13. Let circle: x ^ 2 + y ^ 2 - (2a ^ 2-4) x-4a ^ 2Y + 5A ^ 4-4 = 0, find the value range of real number a and the trajectory equation of center C
- 14. It is known that the square of the equation KX of x minus (3K-1) x + 2 (k-1) = 0. Proof: no matter what the real number k is, the equation always has real number It is known that the square of the equation KX minus (3K-1) x + 2 (k-1) = 0 Proof: no matter what the real number k is, the equation always has real roots (2) If this equation has two real roots, x1, X2, and | x1, X2 | = 2, find the value of K?
- 15. Given the function image y = KX + (3K-1) + 2, where k is a real number, it is proved that no matter what the value of K is, the image of the original function must pass through these two points RT
- 16. Given the circle C: (x-1) 2 + (y + 2) 2 = 9, the straight line L: 2ax-y + 2a-1 = 0, we prove that no matter what the real number a is, the straight line l always intersects the circle
- 17. Given the circle equation x ^ 2 + y ^ 2-2ax-4ay + 5A ^ 2-4 = 0 (a is not equal to 0), we can determine the linear equation that no matter what the value of a is cut by the circle, the chord length is 1
- 18. If a is a real number, then the linear equation of the center of the circle (x-a) 2 + (y + 2a) 2 = 1 is () A. 2x+y=0B. x+2y=0C. x-2y=0D. 2x-y=0
- 19. The moving circle passes through point a (2,0) and is tangent to the square of circle (x + 2) + the square of y = 4. The orbit equation of the center of the moving circle is obtained
- 20. Try to determine the range of real number a so that the positive root of equation x & # 178; - ax + A & # 178; - 4 = 0 has and only has one?