Mathematics Test - 23

\(x=\frac{2 a\left(1-t^{2}\right)}{1+t^{2}}\) and \(y=\frac{4 a t}{1+t^{2}}\)
let, \(t=\tan \theta\)
\(x=2 a\left(\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}\right) \Rightarrow x=2 a \cos 2 \theta\)
\(y=\frac{4 a \tan \theta}{1+\tan ^{2} \theta} \Rightarrow y=2 a \sin 2 \theta\)
\(\therefore \frac{x^{2}}{4 a^{2}}+\frac{y^{2}}{4 a^{2}}=1\)
\(x^{2}+y^{2}=4 a^{2}\)
So, the radius of circle is \(2 a\)
Hence, the correct option is (B)

Put \(\cos \theta=\frac{x}{r} \& \sin \theta=\frac{y}{r}\) in \(r^{2}-4 r(\cos \theta+\sin \theta)-4=0\)
where, \(r=\sqrt{x^{2}+y^{2}}\)
\(x^{2}+y^{2}-4 r\left(\frac{x}{r}+\frac{y}{r}\right)-4=0\)
\(\Rightarrow x^{2}+y^{2}-4 x-4 y-4=0\)
Therefore, center of circle is (2,2)
Hence, the correct option is (C)

since \(a+b=10\) hence the equation of the line can be re-written as
\(\frac{x}{a}+\frac{y}{10-a}=1 \ldots(1)\)
Now the line intersects the axes at \(A=(a, 0)\) and \(B=(0,10-a)\)
The midpoint of \(\mathrm{AB}\) is
\(G=\frac{a+0}{2}, \frac{0+10-a}{2}\)
\(=\left(\frac{a}{2}, \frac{10-a}{2}\right)\)
Hence \((x, y)=\left(\frac{a}{2}, \frac{10-a}{2}\right)\)
Therefore \(x=\frac{a}{2} \ldots\left(\right.\) i) and \(y=\frac{10-a}{2} \ldots\) (ii)
\(y=5-\frac{a}{2}\)
\(y=5-x \ldots(\) from 1\()\)
Hence
\(x+y=5\)
Thus the required locus is \(x+y=5\)
Hence, the correct option is (C)

Let the equation of line be \(\frac{x}{a}+\frac{y}{b}=1\)
where \(a\) and \(b\) are the x-intercept and y-intercept. Then the coordinates of \(\mathrm{A}\) and \(\mathrm{B}\) are \((a, 0)\) and \((0, b)\) Distance of origin to the line is \(c\) \(\Rightarrow c=\frac{|-1|}{\sqrt{\left(\frac{1}{a}\right)^{2}+\left(\frac{1}{b}\right)^{2}}}\)
\(\Rightarrow \frac{1}{c}=\sqrt{\frac{1}{a^{2}}+\frac{1}{b^{2}}}\)
\(\Rightarrow \frac{1}{c^{2}}=\frac{1}{a^{2}}+\frac{1}{b^{2}} \ldots .(1)\)
Let the center of circle through \(O, A, B\) be \((h, k)\) Points \(O(0,0), A(a, 0),(0, b)\) forms a right triangle. So, the center of circle is the mid-point of \(\mathrm{AB}\) i.e. \(\left(\frac{a}{2}, \frac{b}{2}\right)\) \(\Rightarrow h=\frac{a}{2}, k=\frac{b}{2}\)
\(\Rightarrow a=2 h, b=2 k\)
Substitute this value in ( 1 ), we get \(\frac{1}{c^{2}}=\frac{1}{4 h^{2}}+\frac{1}{4 k^{2}}\)
\(\Rightarrow \frac{4}{c^{2}}=\frac{1}{h^{2}}+\frac{1}{k^{2}}\)
\(\Rightarrow 4 c^{-2}=x^{-2}+y^{-2}\) (Replacing \(h, k\) by \(\left.x, y\right)\)
Hence, the correct option is (D)

\(x=1+4 \cos \theta\) and \(y=2+3 \sin \theta\)
Hence \(\frac{x-1}{4}=\cos \theta\) and \(\frac{y-2}{3}=\sin \theta\)
Therefore
\(\cos ^{2} \theta+\sin ^{2} \theta=1\)
Or \(\left(\frac{x-1}{4}\right)^{2}+\left(\frac{y-2}{3}\right)^{2}=1\)
\(\frac{(x-1)^{2}}{16}+\frac{(y-2)^{2}}{9}=1\)
\(\frac{9(x-1)^{2}+16(y-2)^{2}}{144}=1\)
\(O r\)
\(9(x-1)^{2}+16(y-2)^{2}=144\)
Hence, the correct option is (B)

since, the circle touches both negative axes and radius is 3 Center will be (-3,-3) Equation of the circle having centre (-3,3) and radius 3 is given by
\((x+3)^{2}+(y+3)^{2}=(3)^{2}\)
\(\Rightarrow y^{2}+x^{2}+6 x+6 y+9=0\)
Hence, the correct option is (C)

Centre of small circle is at \((r, r) . r\) is a radius of small circle and centre of large circle is at \((R, R) . R\) is a radius of large circle. \(\therefore \sqrt{(R-r)^{2}+(R-r)^{2}}=R+r\)
\(2\left(R^{2}+r^{2}-2 R r\right)=R^{2}+r^{2}+2 r R\)
\(R^{2}+r^{2}-6 R r=0\)
Given \(r=1\) \(R^{2}-6 R+1=0\)
\(R=\frac{6 \pm \sqrt{36-4}}{2}=3 \pm 2 \sqrt{2}\)
Given \(R>r\) So, \(R=3+2 \sqrt{2}\)
Hence, the correct option is (A)

Area of triangle \(=\left|\frac{x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)}{2}\right|\)
The points are (2,3)(-3,4)\((h, k)\) \(\frac{17}{2}=\left|\frac{2(4-k)+(-3)(k-3)+h(3-4)}{2}\right|\)
\(17^{2}=(8-2 k-3 k+9-h)^{2}\)
\(17^{2}=(h+5 k-17)^{2}\)
\(17^{2}=h^{2}+25 k^{2}+17^{2}+10 h k-34 h-170 k\)
\(h^{2}+25 k^{2}+10 h k-34 h-170 k=0\)
Therefore, the locus is \(x^{2}+25 y^{2}+10 y x-34 x-170 y=0\)
Hence, the correct option is (A)

Let \((h, k)\) be any arbitrary point on the locus.
It is given that \((3,4),(-4,3),(h, k)\) are collinear. Hence, area of the triangle formed by these points will be 0 \(\left|\begin{array}{lll}x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1\end{array}\right|=0\)
Condition of col-linearity of three points \(\left(x_{1}, y_{1}\right)\left(x_{2}, y_{2}\right),\left(x_{3}, y_{3}\right)\)
\(\left|\begin{array}{ccc}3 & 4 & 1 \\ -4 & 3 & 1 \\ h & k & 1\end{array}\right|=0\)
\(\therefore h(4-3)-k(4+3)+1(9+16)=0\)
\(\Longrightarrow h-7 k+25=0\)
\(\Longrightarrow x-7 y+25=0\) is the required equation of locus. Hence, option D is correct.
Hence, the correct option is (D)

Let the point be \(P(x, y)\) Then \(\sqrt{(x-(a+b))^{2}+(y-(a-b))^{2}}=\sqrt{(x-(a-b))^{2}+(y-(a+b))^{2}}\)
\_x-(a+b^{2}+(y-(a-b))^{2}=(x-(a-b))^{2}+(y-(a+b))^{2}\)
\_x-(a+b^{2}-(x-(a-b))^{2}=(y-(a+b))^{2}-(y-(a-b))^{2}\)
\_{2 x-2 a)(a-b-(a+b}=(2 y-2 a)(a-b-(a+b))\)
\(2(x-a)(-2 b)=2(y-a)(-2 b)\)
\(x-a=y-a\)
or
\(x-y=0\) is the required equation.
Hence, the correct option is (D)