A) \[0\]
B) \[1\]
C) \[-1\]
D) will depend on\[h\]
Correct Answer: C
Solution :
Key Idea: Equation of tangent at \[({{x}_{1}},\,\,{{y}_{1}})\] the circle\[{{x}^{2}}+{{y}^{2}}={{a}^{2}}\]is\[x{{x}_{1}}+y{{y}_{1}}={{a}^{2}}\]. Given circle \[{{x}^{2}}+{{y}^{2}}={{a}^{2}}\] and point\[(h,\,\,h)\]. \[\therefore \]Equation of tangent at\[(h,\,\,h)\] is \[xh+yh={{a}^{2}}\Rightarrow x+y=\frac{{{a}^{2}}}{h}\] \[\therefore \] \[y=-x+\frac{{{a}^{2}}}{h}\] \[\therefore \]Slope of tangent is\[-1\]. Alternate Method: Given curve \[{{x}^{2}}+{{y}^{2}}={{a}^{2}}\] On differentiating w.r.t.\[x\] we get \[2x+2y\frac{dy}{dx}=0\Rightarrow \frac{dy}{dx}=\frac{-2x}{2y}\] \[\therefore \]Slope of tangent at\[(h,\,\,h)\] is \[{{\left( \frac{dy}{dx} \right)}_{(h,\,\,h)}}=\frac{-2h}{2h}=-1\]
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