已知圆C:x²+y²-4x-14y+45=0及点Q(-2,3)(1)若点P(m,m+1)在圆上,求直线PQ的斜率及直线PQ与圆C的相交弦PE的长度(2)若M(x,y)是圆上任意一点,求(y-3)/(x+2)求PE长!!!!别抄
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已知圆C:x²+y²-4x-14y+45=0及点Q(-2,3)(1)若点P(m,m+1)在圆上,求直线PQ的斜率及直线PQ与圆C的相交弦PE的长度(2)若M(x,y)是圆上任意一点,求(y-3)/(x+2)求PE长!!!!别抄
已知圆C:x²+y²-4x-14y+45=0及点Q(-2,3)
(1)若点P(m,m+1)在圆上,求直线PQ的斜率及直线PQ与圆C的相交弦PE的长度
(2)若M(x,y)是圆上任意一点,求(y-3)/(x+2)
求PE长!!!!
别抄
已知圆C:x²+y²-4x-14y+45=0及点Q(-2,3)(1)若点P(m,m+1)在圆上,求直线PQ的斜率及直线PQ与圆C的相交弦PE的长度(2)若M(x,y)是圆上任意一点,求(y-3)/(x+2)求PE长!!!!别抄
1、(x-2)^2+(y-7)^2=8,
圆心(2,7),半径为2√2,
(m-2)^2+(m+1-7)^2=8,
m^2-8m+16=0,
m=4,
P(4,5),
斜率k=(5-3)/(4+2)=1/3,
PQ方程为:
(y-3)/(x+2)=1/3,
x-3y+11=0,
根据点线距离公式,
圆心至PQ距离d=|2-3*7+11|/√(1+9)
=8/√10=4√10/5.
PE/2=√(R^2-d^2)=√(8-160/25)=2√10/5,
∴PE=4√10/5.
2、k=(y-3)/(x+2)为直线PQ的斜率,
y=kx+2k+3,
kx-y+2k+3=0,
k值在两条切线的斜率之间,
圆心至切线距离,
根据点线距离公式,
|2k-7+2k+3|/√(k^2+1)=R=2√2,
k^2-4k+1=0,
k=(4±2√3)/2
=2±√3,
∴2-√3