Question

Let ABC be a triangle whose vertices are A(-5,5) and B(7, -1). If vertex C lies on the circle withdirector circle has equation x² + y2 = 100, then the locus of orthocenter of AABC is(A) x² + y2 - 4x - By - 30 = 0(B) x² + y2 + 4x - 8y - 20 = 0(C) x² + y2 + 4x + 8y - 30 = 0(D) x² + y2 + 4x + 8y = 0answer is A,but I do not know how?​

Answer 1

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Answer 2

Given: Vertices of the triangle $ABC$are$A(-5,5) and B(7, -1)$ and the equation of the circle is$x^{2} + y^{2} = 100$.

We have to find the locus of the orthocenter of the triangle.

$\text { As we know that he circle whose director circle is } x^{2}+y^{2}=100 \text {, is the circle } x^{2}+y^{2}=50 \text {. }\\\\\mathrm{A}(-5,5) \text { and } \mathrm{B}(7,-1) \text { lie on circle } x^{2}+y^{2}=50 \text {. }(as A,B satistifies the equation)\\\\So, \mathrm{C}(\mathrm{a}, \mathrm{b}) \text { lies on a circle } x^{2}+y^{2}=50\\\\A(-5,5), B(7,-1) C(a, b) \text { lie on a circle } x^{2}+y^{2}=50 \text {. }\\\\\therefore \quad \text { C satisfies } a^{2}+b^{2}=50$

$\therefore x^{2}+y^{2}=50 \text { is a circum circle of } \triangle A B C \text {. Circum center }=(0,0) \text {. }\\\mathrm{O}=\text { Circumcenter }=(0,0)\\\\\mathrm{G}=\text { Centroid }=\left(\frac{-5+7+a}{3}, \frac{5-1+b}{3}\right)=\left(\frac{2+a}{3}, \frac{4+b}{3}\right)\\\\\text { Let } H=\text { orthocenter }=(x, y)\\\\\text { We know that } G \text { divides the join of HO in the ration } 2: 1 \text { internally. }$

$\text { Therefore by section formula: } x=a+2 \text { and } y=b+4 \text {, }\\\text { we get the required locus of orthocenter as: }\\(x-2)^{2}+(y-4)^{2}=50$

$=>x^{2}+y^{2}-4 x-8 y=30$

Hence, the locus of the orthocenter is $x^{2}+y^{2}-4 x-8 y=30$