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QJEEM24Ma72c7: JEE Main 2024 Mathematics - Class 11 JEEM24Ma72c7 Solution: If be the orthocentre of the triangle whose vertices are and ,…

If P(6,1)\mathrm{P}(6,1) be the orthocentre of the triangle whose vertices are A(5,−2),B(8,3)\mathrm{A}(5,-2), \mathrm{B}(8,3) and C(h,k)\mathrm{C}(\mathrm{h}, \mathrm{k}) , then the point C\mathrm{C} lies on the circle :

Multiple Choice Options

  • A x2+y2−74=0x^2+y^2-74=0
  • B x2+y2−65=0x^2+y^2-65=0
  • C x2+y2−61=0x^2+y^2-61=0
  • D x2+y2−52=0x^2+y^2-52=0

Step-by-Step Verified Solution

JEE Main 2024 (Online) 6th April Evening Shift Mathematics - Circle Question 8 English Explanation

Slope of BF=(mBF)=3−18−6=1B F=\left(m_{B F}\right)=\frac{3-1}{8-6}=1

amp;mAC=k+2h−5=−1(AC⊥BF)amp;⇒k+2=5−hamp;⇒k=3−h… (i) amp;mAD=1+26−5=3amp;mBC=k−3h−8=−13(AD⊥BC)amp;⇒3k−9=8−h… (ii) \begin{aligned}& m_{A C}=\frac{k+2}{h-5}=-1(A C \perp B F) \\& \Rightarrow k+2=5-h \\& \Rightarrow k=3-h \quad \ldots \text { (i) } \\& m_{A D}=\frac{1+2}{6-5}=3 \\& m_{B C}=\frac{k-3}{h-8}=-\frac{1}{3}(A D \perp B C) \\& \Rightarrow 3 k-9=8-h \quad \ldots \text { (ii) }\end{aligned}

From (i) & (ii)

h=−4,k=7h=-4, k=7

Now, h2+k2=16+49=65h^2+k^2=16+49=65

h2+k2−65=0h^2+k^2-65=0

Locus is x2+y2−65=0x^2+y^2-65=0

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