Q.23. Let the line intersect the circle C : at the points Q and R. If P(, ) is a point on C such t

Q: What is the correct answer for this exam Mathematics Q23?

The correct answer is 18. The circle C has center $(4, -3)$ and radius $r=3$.The line is $x-y=4$.This line passes through the center of the circle, because $4 - (-3) = 7 \ne 4$. So the line is not passing through the center.The line $x-y=4$ can be written as $y=x-4$.Substitute $y=x-4$ into the circle equation:$(x-4)^2 + ((x-4)+3)^2 = 9$.$(x-4)^2 + (x-1)^2 = 9$.$x^2-8x+16 + x^2-2x+1 = 9$.$2x^2 - 10x + 17 = 9$.$2x^2 - 10x +

Q: Q.23. Let the line intersect the circle C : at the points Q and R. If P(, ) is a point on C such that PQ = PR, then is equal to

The correct answer is 18. The circle C has center $(4, -3)$ and radius $r=3$.The line is $x-y=4$.This line passes through the center of the circle, because $4 - (-3) = 7 \ne 4$. So the line is not passing through the center.The line $x-y=4$ can be written as $y=x-4$.Substitute $y=x-4$ into the circle equation:$(x-4)^2 + ((x-4)+3)^2 = 9$.$(x-4)^2 + (x-1)^2 = 9$.$x^2-8x+16 + x^2-2x+1 = 9$.$2x^2 - 10x + 17 = 9$.$2x^2 - 10x +

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Q.23. Let the line  intersect the circle C :  at the points Q and R. If P(, ) is a point on C such that PQ = PR, then  is equal to

FirstInTest · Accepted Answer

The circle C has center $(4, -3)$ and radius $r=3$.The line is $x-y=4$.This line passes through the center of the circle, because $4 - (-3) = 7 
e 4$. So the line is not passing through the center.The line $x-y=4$ can be written as $y=x-4$.Substitute $y=x-4$ into the circle equation:$(x-4)^2 + ((x-4)+3)^2 = 9$.$(x-4)^2 + (x-1)^2 = 9$.$x^2-8x+16 + x^2-2x+1 = 9$.$2x^2 - 10x + 17 = 9$.$2x^2 - 10x + 8 = 0$.$x^2 - 5x + 4 = 0$.$(x-1)(x-4) = 0$.So the intersection points are $x=1$ and $x=4$.If $x=1$, $y=1-4=-3$. So $Q=(1,-3)$.If $x=4$, $y=4-4=0$. So $R=(4,0)$.P($\alpha$, $\beta$) is a point on C such that PQ = PR.This means P is equidistant from Q and R.The locus of points equidistant from Q and R is the perpendicular bisector of the segment QR.Since P is on the circle and equidistant from Q and R, P must be one of the endpoints of the diameter that is perpendicular to the chord QR.The midpoint of QR is $M = (\frac{1+4}{2}, \frac{-3+0}{2}) = (\frac{5}{2}, -\frac{3}{2})$.The slope of QR is $m_{QR} = \frac{0 - (-3)}{4 - 1} = \frac{3}{3} = 1$.The slope of the perpendicular bisector is $m_{\perp} = -1$.The equation of the perpendicular bisector is $y - (-\frac{3}{2}) = -1(x - \frac{5}{2})$.$y + \frac{3}{2} = -x + \frac{5}{2}$.$y = -x + \frac{5}{2} - \frac{3}{2} = -x + \frac{2}{2} = -x+1$.So $y=-x+1$.The point P($\alpha$, $\beta$) lies on this line, so $\beta = -\alpha+1$.Also, P($\alpha$, $\beta$) lies on the circle C: $(\alpha-4)^2 + (\beta+3)^2 = 9$.Substitute $\beta = -\alpha+1$:$(\alpha-4)^2 + ((- \alpha+1)+3)^2 = 9$.$(\alpha-4)^2 + (4-\alpha)^2 = 9$.$2(\alpha-4)^2 = 9$.$(\alpha-4)^2 = \frac{9}{2}$.$\alpha-4 = \pm \frac{3}{\sqrt{2}} = \pm \frac{3\sqrt{2}}{2}$.$\alpha = 4 \pm \frac{3\sqrt{2}}{2}$.If $\alpha = 4 + \frac{3\sqrt{2}}{2}$, then $\beta = - (4 + \frac{3\sqrt{2}}{2}) + 1 = -3 - \frac{3\sqrt{2}}{2}$.If $\alpha = 4 - \frac{3\sqrt{2}}{2}$, then $\beta = - (4 - \frac{3\sqrt{2}}{2}) + 1 = -3 + \frac{3\sqrt{2}}{2}$.We need to calculate $(6\alpha + 8\beta)^2$.Case 1: $\alpha = 4 + \frac{3\sqrt{2}}{2}$, $\beta = -3 - \frac{3\sqrt{2}}{2}$.$6\alpha + 8\beta = 6(4 + \frac{3\sqrt{2}}{2}) + 8(-3 - \frac{3\sqrt{2}}{2})$$= 24 + 9\sqrt{2} - 24 - 12\sqrt{2} = -3\sqrt{2}$.$(6\alpha + 8\beta)^2 = (-3\sqrt{2})^2 = 9 	imes 2 = 18$.Case 2: $\alpha = 4 - \frac{3\sqrt{2}}{2}$, $\beta = -3 + \frac{3\sqrt{2}}{2}$.$6\alpha + 8\beta = 6(4 - \frac{3\sqrt{2}}{2}) + 8(-3 + \frac{3\sqrt{2}}{2})$$= 24 - 9\sqrt{2} - 24 + 12\sqrt{2} = 3\sqrt{2}$.$(6\alpha + 8\beta)^2 = (3\sqrt{2})^2 = 9 	imes 2 = 18$.In both cases, the value is 18.The solution in the image for Q.23 is 18. It uses a different approach: P($\alpha$, $\beta$) is on the circle and PQ=PR. This means P is the point on the circle such that the tangent at P is parallel to QR, or P is the point on the circle such that the line CP is perpendicular to QR.The center of the circle is $C(4,-3)$. The slope of QR is 1.The line CP must have slope $-1$.Equation of line CP: $y - (-3) = -1(x - 4) \Rightarrow y+3 = -x+4 \Rightarrow y = -x+1$.This is the same line $y=-x+1$ we found as the perpendicular bisector.So P($\alpha$, $\beta$) is the intersection of $y=-x+1$ and the circle. This is what we calculated.The solution in the image uses $(PQ)^2 = (PR)^2 \Rightarrow (\alpha-4)^2 + \beta^2 = (\alpha-1)^2 + (\beta+3)^2$.This is incorrect. The points Q and R are $(1,-3)$ and $(4,0)$.So $(\alpha-1)^2 + (\beta-(-3))^2 = (\alpha-4)^2 + (\beta-0)^2$.$(\alpha-1)^2 + (\beta+3)^2 = (\alpha-4)^2 + \beta^2$.$\alpha^2-2\alpha+1 + \beta^2+6\beta+9 = \alpha^2-8\alpha+16 + \beta^2$.$-2\alpha+1+6\beta+9 = -8\alpha+16$.$-2\alpha+6\beta+10 = -8\alpha+16$.$6\alpha+6\beta = 6$.$\alpha+\beta = 1$.This is the equation of the perpendicular bisector of QR. This is correct.My equation for the perpendicular bisector was $y=-x+1$, which is $x+y=1$. So $\alpha+\beta=1$.This is consistent.Then the calculation $6\alpha+8\beta = 6\alpha+8(1-\alpha) = 6\alpha+8-8\alpha = 8-2\alpha$.We also have $(\alpha-4)^2 + (\beta+3)^2 = 9$.Substitute $\beta=1-\alpha$:$(\alpha-4)^2 + ((1-\alpha)+3)^2 = 9$.$(\alpha-4)^2 + (4-\alpha)^2 = 9$.$2(\alpha-4)^2 = 9$.$(\alpha-4)^2 = \frac{9}{2}$.$\alpha-4 = \pm \frac{3}{\sqrt{2}}$.$\alpha = 4 \pm \frac{3}{\sqrt{2}}$.If $\alpha = 4 + \frac{3}{\sqrt{2}}$, then $8-2\alpha = 8 - 2(4 + \frac{3}{\sqrt{2}}) = 8 - 8 - \frac{6}{\sqrt{2}} = -3\sqrt{2}$.Square is $(-3\sqrt{2})^2 = 18$.If $\alpha = 4 - \frac{3}{\sqrt{2}}$, then $8-2\alpha = 8 - 2(4 - \frac{3}{\sqrt{2}}) = 8 - 8 + \frac{6}{\sqrt{2}} = 3\sqrt{2}$.Square is $(3\sqrt{2})^2 = 18$.The result is consistent. The solution in the image is correct.

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