Question

Point P(-10,14) is a point external to a circle. Points A(5,6), B(x,y), C(1,4) are on the circle. Line PC is tangent to circle and P, A, and B are collinear, with B between P and A. Find the coordinates of point B. Give your answers two decimal places?

Answer 1

`B(x,y)=(25/17,134/17)=(1.47,7.88)`

Explanation:

By Tangent-Secant theorem, we know that,
`PC^2=PA*PB`
`PC^2=(4-14)^2+(1+10)^2=100+121=221`,
`PA=sqrt((6-14)^2+(5+10)^2)=sqrt(64+225)=sqrt289=17`,
`=> PB=(PC^2)/(PA)=221/17=13`
`=> PB:BA=13:(17-13)=13:4`
`=> B` divides line `PA` in the ratio of `13:4`
Section formula : if a point `B(x,y)` divides a line joining `P(x_1,y_1) and A(x_2,y_2)` in the ratio of `m:n`,
then `B(x,y) = ((m*x_2+n*x_1)/(m+n), (m*y_2+n*y_1)/(m+n))`

`=> B(x,y)=((13*5+4*(-10))/(13+4), (13*6+4*14)/(13+4))`
`=(25/17,134/17)=(1.47,7.88)`