find the equation of the tangent at (0,2) to the circle with equation(x+2)平方 + (y+1)平方=13

来源:学生作业帮助网 编辑:作业帮 时间:2024/05/05 10:15:02
find the equation of the tangent at (0,2) to the circle with equation(x+2)平方 + (y+1)平方=13

find the equation of the tangent at (0,2) to the circle with equation(x+2)平方 + (y+1)平方=13
find the equation of the tangent at (0,2) to the circle with equation
(x+2)平方 + (y+1)平方=13

find the equation of the tangent at (0,2) to the circle with equation(x+2)平方 + (y+1)平方=13
(x+2)平方 + (y+1)平方=13的圆心为(-2,-1),所以过圆心和(0,2)的直线斜率为:[2-(-1)]/(0-(-2))=3/2,所以切线方程显然与过圆心和(0,2)的直线斜率互为负倒数:所以为:-1/(3/2)=-2/3,已知斜率和过(0,2)点的条件,可得:y=(-2/3)x+2,整理得:2x+3y=6

找到过点(0,2)的圆的切线方程
答案
2x+3y=6
因为该点在圆上,所以过这点只有一条切线
解法:
代入点可知:2*2+3*3=13
知点在圆上,所以这点既在圆上又在切线上
所以有
(2+0)*(2+x)+(2+1)*(y+1)=13
整理得:2x+3y=6
即切线方程...

全部展开

找到过点(0,2)的圆的切线方程
答案
2x+3y=6
因为该点在圆上,所以过这点只有一条切线
解法:
代入点可知:2*2+3*3=13
知点在圆上,所以这点既在圆上又在切线上
所以有
(2+0)*(2+x)+(2+1)*(y+1)=13
整理得:2x+3y=6
即切线方程

收起

题目意思为:求过圆上一点(0,2)的切线方程。
设此点为A(0,2),圆心O坐标为(-2,-1),则直线OA斜率k=(-1-2)/(-2)=3/2.
设所求切线L斜率为k′,由相切关系,得直线OA垂直于直线L,故有斜率之积k*k′=-1,
从而得k′=-2/3.
所求切线方程为:y-2=-2/3x,即 y=-2/3x+2....

全部展开

题目意思为:求过圆上一点(0,2)的切线方程。
设此点为A(0,2),圆心O坐标为(-2,-1),则直线OA斜率k=(-1-2)/(-2)=3/2.
设所求切线L斜率为k′,由相切关系,得直线OA垂直于直线L,故有斜率之积k*k′=-1,
从而得k′=-2/3.
所求切线方程为:y-2=-2/3x,即 y=-2/3x+2.

收起

2x+3y-6=0

依题求点(0,2)处的切线方程,设圆心为O,并有A(0,2)
易得直线OA的方程为y=3/2x+2,由于切线垂直于OA
则切线斜率为-2/3,即y=-2/3x+b,代入A(0,2)得b=2
即该圆O在点(0,2)处的切线方程为y=-2/3x+2

Find the equation of the line OP /> 这么做这数学(find the equation of the tangent) Find all of the solutions of th equation cosx+cos2x=0 Find all of the solutions of th equation cosx+cos2x=0 Graph and find the slope of the following equation :Y = 100 – 20 X find the equation of the tangent at (0,2) to the circle with equation(x+2)平方 + (y+1)平方=13 Find the equation of line tangent to the graph of f(x)=e+ln x at x=1? Find the equation of the perpendicular bisector of A(-2,1)and B(4,-5)perpendicular垂直bisector 等分 find the equation of the normal at the point (1,2)to the graph y=x+1/x Find the equation of the tangent to the curve y=(x^2-5)^3 at the point (2,-1) Aand B are the roots of the equation x2-4x+6=0(i) Find the value of 1 1--- + ---A B 1 1(ii) Find the quadratic equation whose roots are -- and --A B 帮我做2道英文数学题 谢谢1.Find the equation of the line which passes through the point (6.-4).And is parallel to the line 2x-3y+3=0.2.Find the equation of the line which passes through the points P(0,A) and Q(A,2A) .3.Find the equation of urgent!if x satifies the equation sinx+cosx =√2,find the value of tan (x/2)with process find an equation for a line that is normal to the graph of y = xe^x and goes Find the equation of the locus of the mid-point of the chord ofthe curve x²+4y²=1 with gradient 1. Find an equation of the tangent line to the graph of the function f(x) = 8/(√x^2+3x) at the point (1,4) for 的用法是什么句型Find a Cartesian equation for the path of the falling cargo 为什么不用of呢?for 和of 的区别呢 1,P is the point (7,5) and L1 is the line with equation 3x+4y=16(1),Find out the equation of the line L2 which passes through P and is perpendicular to L2.(2),Find the point of intersection of the L1 and L2.(3),Find the perpendicular distance of P fr