Let \(P=(0,2)\). Let \(A(x_1,y_1)\) and \(B(x_2,y_2)\) be two points on the circle \(x^2+y^2-6x+4y+4=0\) such that \(PA\) is minimum and \(PB\) is maximum. Then \(\frac{3(y_1-y_2){(x_2-x_1)}=\)}
Show Hint
- \(8\)
- \(2\)
- \(1\)
- \(4\)
The Correct Option is D
Solution and Explanation
\[ x^2+y^2-6x+4y+4=0 \] \[ (x-3)^2+(y+2)^2=9 \] Center: \[ (3,-2) \] Radius: \[ 3 \]
Step 2: Use nearest and farthest points.
Point: \[ P=(0,2) \] Vector from center to \(P\): \[ (-3,4) \] Magnitude: \[ 5 \] Nearest point: \[ A=\left(3-\frac{9}{5},-2+\frac{12}{5}\right) \] \[ =\left(\frac65,\frac25\right) \] Farthest point: \[ B=\left(3+\frac95,-2-\frac{12}{5}\right) \] \[ =\left(\frac{24}{5},-\frac{22}{5}\right) \]
Step 3: Substitute.
\[ \frac{3(y_1-y_2)}{x_2-x_1} \] \[ =\frac{3\left(\frac25+\frac{22}{5}\right)}{\frac{24}{5}-\frac65} \] \[ =\frac{3\cdot\frac{24}{5}}{\frac{18}{5}} \] \[ =\frac{72}{18} \] \[ =4 \]
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