Question 39
NECO
•
2019
•
FURTHER_MATHEMATICS
•
objective
Find the equations of all circles that (i) pass through the point \((2,3)\), (ii) are tangent to the line \(3x-4y+5=0\), and (iii) have their centre on the line \(y=x+1\).
Answer Options
A.
\[(x-\frac{99+5\sqrt{2}}{49})^{2}+(y-\frac{148+5\sqrt{2}}{49})^{2}=\frac{102-20\sqrt{2}}{2401}\]\n\[(x-\frac{99-5\sqrt{2}}{49})^{2}+(y-\frac{148-5\sqrt{2}}{49})^{2}=\frac{102+20\sqrt{2}}{2401}\]
Correct Answer
B.
\[(x-\frac{99+5\sqrt{2}}{49})^{2}+(y-\frac{148+5\sqrt{2}}{49})^{2}=\frac{102+20\sqrt{2}}{2401}\]\n\[(x-\frac{99-5\sqrt{2}}{49})^{2}+(y-\frac{148-5\sqrt{2}}{49})^{2}=\frac{102-20\sqrt{2}}{2401}\]
C.
\[(x-\frac{99+4\sqrt{2}}{49})^{2}+(y-\frac{148+4\sqrt{2}}{49})^{2}=\frac{102+16\sqrt{2}}{2401}\]\n\[(x-\frac{99-4\sqrt{2}}{49})^{2}+(y-\frac{148-4\sqrt{2}}{49})^{2}=\frac{102-16\sqrt{2}}{2401}\]
D.
\[(x-\frac{99+5\sqrt{2}}{49})^{2}+(y-\frac{148+5\sqrt{2}}{49})^{2}=\frac{102+20\sqrt{2}}{49}\]\n\[(x-\frac{99-5\sqrt{2}}{49})^{2}+(y-\frac{148-5\sqrt{2}}{49})^{2}=\frac{102-20\sqrt{2}}{49}\]
Correct Answer
B
Correct Option:
\[(x-\frac{99+5\sqrt{2}}{49})^{2}+(y-\frac{148+5\sqrt{2}}{49})^{2}=\frac{102+20\sqrt{2}}{2401}\]\n\[(x-\frac{99-5\sqrt{2}}{49})^{2}+(y-\frac{148-5\sqrt{2}}{49})^{2}=\frac{102-20\sqrt{2}}{2401}\]
About This Question
This is
Neco
2019
Further Mathematics
Question 39.
It is one of the
objective
questions from the
2019
Neco
Further Mathematics
examination.
Difficulty level:
Medium
.