119761
For how many values of \(k\), the line \(3 x-4 y=\lambda\) may touch the circle \(x^2+y^2-4 x-8 y-5=0\) ?
1 1
2 2
3 3
4 None of the values of \(\mathrm{k}\)
Explanation:
B The line \(3 \mathrm{x}-4 \mathrm{y}=\lambda \text { touches the circle }\) \(\quad \mathrm{x}^2+\mathrm{y}^2-4 \mathrm{x}-8 \mathrm{y}-5=0\) \(\quad \mathrm{C}(-\mathrm{g},-\mathrm{f})=(2,4)\) \(=\sqrt{\mathrm{g}^2+\mathrm{f}^2-\mathrm{c}}=\sqrt{4+16+5}=5\) \(\text { We know distance of a straight line } \mathrm{ax}+\mathrm{by}+\mathrm{c}=0\) \(\mathrm{ax}+\mathrm{by}+\mathrm{c}=0 \text { from any point }\left(\mathrm{x}_1, \mathrm{y}_1\right) \text { is }\) \(\mathrm{d}=\frac{\mathrm{ax} \mathrm{x}_1+\mathrm{by}_1+\mathrm{c}}{\sqrt{\mathrm{a}^2+\mathrm{b}^2}}\) \(5=\frac{\mathrm{ax}_1+\mathrm{by}_1+\mathrm{c}}{\sqrt{\mathrm{a}^2+\mathrm{b}^2}}=\frac{|6-16-\lambda|}{\sqrt{16+9}}=\frac{|-\lambda-10|}{5}\) \(\frac{-\lambda-10}{5}= \pm 5\) \(-\lambda-10= \pm 25= \pm 25-10=-35,15\) We know distance of a straight line \(a x+b y+c=0\) \(a x+b y+c=0\) from any point \(\left(x_1, y_1\right)\) is
SCRA-2015
Conic Section
119762
The line segment joining the foci of the hyperbola \(x^2-y^2+1=0\) is one of the diameters of a circle. The equation of the circle is
1 \(x^2+y^2=4\)
2 \(x^2+y^2=\sqrt{2}\)
3 \(x^2+y^2=2\)
4 \(x^2+y^2=2 \sqrt{2}\)
Explanation:
C Given, equation of hyperbola is \(x^2-y^2+1=0\) \(\Rightarrow \quad \mathrm{y}^2-\mathrm{x}^2=1\) On comparing it with \(\frac{y^2}{b^2}-\frac{x^2}{a^2}=1\). we get \(\quad \mathrm{a}=\mathrm{b}=1\) \(\text { now, } \mathrm{e}=\sqrt{1+\frac{\mathrm{a}^2}{\mathrm{~b}^2}}=\sqrt{1+\frac{1}{1}}=\sqrt{2}\) \(\therefore \text { foci }=(0, \pm \mathrm{be})=(0, \pm \sqrt{2})\) Since, line joining foci of hyperbola is diameter of circle. \(\therefore \text { center of circle }=\left(\frac{0+0}{2}, \frac{\sqrt{2}-\sqrt{2}}{2}\right)=(0,0)\) \(\text { and radius }=\frac{1}{2} \sqrt{(0-0)^2+(\sqrt{2}-(-\sqrt{2}))^2}\) \(\quad=\frac{1}{2} \sqrt{(2 \sqrt{2})^2}=\sqrt{2}\) \(\therefore\) Equation of circle will be \((\mathrm{x}-0)^2+(\mathrm{y}-0)^2=(\sqrt{2})^2\) \(\Rightarrow \quad \mathrm{x}^2+\mathrm{y}^2=2\)
WB JEE-2017
Conic Section
119763
If \(P(0,0), Q(1,0)\) and \(R\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)\) are three given points, then the centre of the circle for which the lines \(P Q, Q R\) and \(R P\) are the tangents is
119764
A variable circle passes through the fixed point \(A(p, q)\) and touches \(x\)-axis. The locus of the other end of the diameter through \(A\) is
1 \((x-p)^2=4 q y\)
2 \((x-q)^2=4 p y\)
3 \((\mathrm{y}-\mathrm{p})^2=4 \mathrm{qx}\)
4 \((y-q)^2=4 \mathrm{px}\)
Explanation:
A Let, ( \(\mathrm{h}, \mathrm{k}\) ) be the coordinates of the other end \(\mathrm{B}\) of the diameter through \(\mathrm{A}\). The coordinates of the centre are \(\left(\frac{\mathrm{p}+\mathrm{h}}{2}, \frac{\mathrm{q}+\mathrm{k}}{2}\right)\) Since, the circle touches \(\mathrm{x}\)-axis Therefore, \(\mid \mathrm{y} \text {-coordinates of its centre } \mid=\text { Radius }\) \(\frac{\mathrm{q}+\mathrm{k}}{2}=\frac{1}{2} \sqrt{(\mathrm{p}-\mathrm{h})^2+(\mathrm{q}-\mathrm{k})^2}\) \((\mathrm{q}+\mathrm{k})^2=(\mathrm{p}-\mathrm{h})^2+(\mathrm{q}-\mathrm{k})^2\) \(\mathrm{q}^2+\mathrm{k}^2+2 \mathrm{qk}=\mathrm{p}^2+\mathrm{q}^2-2 \mathrm{ph}+\mathrm{q}^2+\mathrm{k}^2-2 \mathrm{qk}\) \(4 \mathrm{q}=(\mathrm{h}-\mathrm{p})^2\) Hence, The locus of \((h, k)\) is \((x-p)^2=4 q y\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Conic Section
119761
For how many values of \(k\), the line \(3 x-4 y=\lambda\) may touch the circle \(x^2+y^2-4 x-8 y-5=0\) ?
1 1
2 2
3 3
4 None of the values of \(\mathrm{k}\)
Explanation:
B The line \(3 \mathrm{x}-4 \mathrm{y}=\lambda \text { touches the circle }\) \(\quad \mathrm{x}^2+\mathrm{y}^2-4 \mathrm{x}-8 \mathrm{y}-5=0\) \(\quad \mathrm{C}(-\mathrm{g},-\mathrm{f})=(2,4)\) \(=\sqrt{\mathrm{g}^2+\mathrm{f}^2-\mathrm{c}}=\sqrt{4+16+5}=5\) \(\text { We know distance of a straight line } \mathrm{ax}+\mathrm{by}+\mathrm{c}=0\) \(\mathrm{ax}+\mathrm{by}+\mathrm{c}=0 \text { from any point }\left(\mathrm{x}_1, \mathrm{y}_1\right) \text { is }\) \(\mathrm{d}=\frac{\mathrm{ax} \mathrm{x}_1+\mathrm{by}_1+\mathrm{c}}{\sqrt{\mathrm{a}^2+\mathrm{b}^2}}\) \(5=\frac{\mathrm{ax}_1+\mathrm{by}_1+\mathrm{c}}{\sqrt{\mathrm{a}^2+\mathrm{b}^2}}=\frac{|6-16-\lambda|}{\sqrt{16+9}}=\frac{|-\lambda-10|}{5}\) \(\frac{-\lambda-10}{5}= \pm 5\) \(-\lambda-10= \pm 25= \pm 25-10=-35,15\) We know distance of a straight line \(a x+b y+c=0\) \(a x+b y+c=0\) from any point \(\left(x_1, y_1\right)\) is
SCRA-2015
Conic Section
119762
The line segment joining the foci of the hyperbola \(x^2-y^2+1=0\) is one of the diameters of a circle. The equation of the circle is
1 \(x^2+y^2=4\)
2 \(x^2+y^2=\sqrt{2}\)
3 \(x^2+y^2=2\)
4 \(x^2+y^2=2 \sqrt{2}\)
Explanation:
C Given, equation of hyperbola is \(x^2-y^2+1=0\) \(\Rightarrow \quad \mathrm{y}^2-\mathrm{x}^2=1\) On comparing it with \(\frac{y^2}{b^2}-\frac{x^2}{a^2}=1\). we get \(\quad \mathrm{a}=\mathrm{b}=1\) \(\text { now, } \mathrm{e}=\sqrt{1+\frac{\mathrm{a}^2}{\mathrm{~b}^2}}=\sqrt{1+\frac{1}{1}}=\sqrt{2}\) \(\therefore \text { foci }=(0, \pm \mathrm{be})=(0, \pm \sqrt{2})\) Since, line joining foci of hyperbola is diameter of circle. \(\therefore \text { center of circle }=\left(\frac{0+0}{2}, \frac{\sqrt{2}-\sqrt{2}}{2}\right)=(0,0)\) \(\text { and radius }=\frac{1}{2} \sqrt{(0-0)^2+(\sqrt{2}-(-\sqrt{2}))^2}\) \(\quad=\frac{1}{2} \sqrt{(2 \sqrt{2})^2}=\sqrt{2}\) \(\therefore\) Equation of circle will be \((\mathrm{x}-0)^2+(\mathrm{y}-0)^2=(\sqrt{2})^2\) \(\Rightarrow \quad \mathrm{x}^2+\mathrm{y}^2=2\)
WB JEE-2017
Conic Section
119763
If \(P(0,0), Q(1,0)\) and \(R\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)\) are three given points, then the centre of the circle for which the lines \(P Q, Q R\) and \(R P\) are the tangents is
119764
A variable circle passes through the fixed point \(A(p, q)\) and touches \(x\)-axis. The locus of the other end of the diameter through \(A\) is
1 \((x-p)^2=4 q y\)
2 \((x-q)^2=4 p y\)
3 \((\mathrm{y}-\mathrm{p})^2=4 \mathrm{qx}\)
4 \((y-q)^2=4 \mathrm{px}\)
Explanation:
A Let, ( \(\mathrm{h}, \mathrm{k}\) ) be the coordinates of the other end \(\mathrm{B}\) of the diameter through \(\mathrm{A}\). The coordinates of the centre are \(\left(\frac{\mathrm{p}+\mathrm{h}}{2}, \frac{\mathrm{q}+\mathrm{k}}{2}\right)\) Since, the circle touches \(\mathrm{x}\)-axis Therefore, \(\mid \mathrm{y} \text {-coordinates of its centre } \mid=\text { Radius }\) \(\frac{\mathrm{q}+\mathrm{k}}{2}=\frac{1}{2} \sqrt{(\mathrm{p}-\mathrm{h})^2+(\mathrm{q}-\mathrm{k})^2}\) \((\mathrm{q}+\mathrm{k})^2=(\mathrm{p}-\mathrm{h})^2+(\mathrm{q}-\mathrm{k})^2\) \(\mathrm{q}^2+\mathrm{k}^2+2 \mathrm{qk}=\mathrm{p}^2+\mathrm{q}^2-2 \mathrm{ph}+\mathrm{q}^2+\mathrm{k}^2-2 \mathrm{qk}\) \(4 \mathrm{q}=(\mathrm{h}-\mathrm{p})^2\) Hence, The locus of \((h, k)\) is \((x-p)^2=4 q y\)
119761
For how many values of \(k\), the line \(3 x-4 y=\lambda\) may touch the circle \(x^2+y^2-4 x-8 y-5=0\) ?
1 1
2 2
3 3
4 None of the values of \(\mathrm{k}\)
Explanation:
B The line \(3 \mathrm{x}-4 \mathrm{y}=\lambda \text { touches the circle }\) \(\quad \mathrm{x}^2+\mathrm{y}^2-4 \mathrm{x}-8 \mathrm{y}-5=0\) \(\quad \mathrm{C}(-\mathrm{g},-\mathrm{f})=(2,4)\) \(=\sqrt{\mathrm{g}^2+\mathrm{f}^2-\mathrm{c}}=\sqrt{4+16+5}=5\) \(\text { We know distance of a straight line } \mathrm{ax}+\mathrm{by}+\mathrm{c}=0\) \(\mathrm{ax}+\mathrm{by}+\mathrm{c}=0 \text { from any point }\left(\mathrm{x}_1, \mathrm{y}_1\right) \text { is }\) \(\mathrm{d}=\frac{\mathrm{ax} \mathrm{x}_1+\mathrm{by}_1+\mathrm{c}}{\sqrt{\mathrm{a}^2+\mathrm{b}^2}}\) \(5=\frac{\mathrm{ax}_1+\mathrm{by}_1+\mathrm{c}}{\sqrt{\mathrm{a}^2+\mathrm{b}^2}}=\frac{|6-16-\lambda|}{\sqrt{16+9}}=\frac{|-\lambda-10|}{5}\) \(\frac{-\lambda-10}{5}= \pm 5\) \(-\lambda-10= \pm 25= \pm 25-10=-35,15\) We know distance of a straight line \(a x+b y+c=0\) \(a x+b y+c=0\) from any point \(\left(x_1, y_1\right)\) is
SCRA-2015
Conic Section
119762
The line segment joining the foci of the hyperbola \(x^2-y^2+1=0\) is one of the diameters of a circle. The equation of the circle is
1 \(x^2+y^2=4\)
2 \(x^2+y^2=\sqrt{2}\)
3 \(x^2+y^2=2\)
4 \(x^2+y^2=2 \sqrt{2}\)
Explanation:
C Given, equation of hyperbola is \(x^2-y^2+1=0\) \(\Rightarrow \quad \mathrm{y}^2-\mathrm{x}^2=1\) On comparing it with \(\frac{y^2}{b^2}-\frac{x^2}{a^2}=1\). we get \(\quad \mathrm{a}=\mathrm{b}=1\) \(\text { now, } \mathrm{e}=\sqrt{1+\frac{\mathrm{a}^2}{\mathrm{~b}^2}}=\sqrt{1+\frac{1}{1}}=\sqrt{2}\) \(\therefore \text { foci }=(0, \pm \mathrm{be})=(0, \pm \sqrt{2})\) Since, line joining foci of hyperbola is diameter of circle. \(\therefore \text { center of circle }=\left(\frac{0+0}{2}, \frac{\sqrt{2}-\sqrt{2}}{2}\right)=(0,0)\) \(\text { and radius }=\frac{1}{2} \sqrt{(0-0)^2+(\sqrt{2}-(-\sqrt{2}))^2}\) \(\quad=\frac{1}{2} \sqrt{(2 \sqrt{2})^2}=\sqrt{2}\) \(\therefore\) Equation of circle will be \((\mathrm{x}-0)^2+(\mathrm{y}-0)^2=(\sqrt{2})^2\) \(\Rightarrow \quad \mathrm{x}^2+\mathrm{y}^2=2\)
WB JEE-2017
Conic Section
119763
If \(P(0,0), Q(1,0)\) and \(R\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)\) are three given points, then the centre of the circle for which the lines \(P Q, Q R\) and \(R P\) are the tangents is
119764
A variable circle passes through the fixed point \(A(p, q)\) and touches \(x\)-axis. The locus of the other end of the diameter through \(A\) is
1 \((x-p)^2=4 q y\)
2 \((x-q)^2=4 p y\)
3 \((\mathrm{y}-\mathrm{p})^2=4 \mathrm{qx}\)
4 \((y-q)^2=4 \mathrm{px}\)
Explanation:
A Let, ( \(\mathrm{h}, \mathrm{k}\) ) be the coordinates of the other end \(\mathrm{B}\) of the diameter through \(\mathrm{A}\). The coordinates of the centre are \(\left(\frac{\mathrm{p}+\mathrm{h}}{2}, \frac{\mathrm{q}+\mathrm{k}}{2}\right)\) Since, the circle touches \(\mathrm{x}\)-axis Therefore, \(\mid \mathrm{y} \text {-coordinates of its centre } \mid=\text { Radius }\) \(\frac{\mathrm{q}+\mathrm{k}}{2}=\frac{1}{2} \sqrt{(\mathrm{p}-\mathrm{h})^2+(\mathrm{q}-\mathrm{k})^2}\) \((\mathrm{q}+\mathrm{k})^2=(\mathrm{p}-\mathrm{h})^2+(\mathrm{q}-\mathrm{k})^2\) \(\mathrm{q}^2+\mathrm{k}^2+2 \mathrm{qk}=\mathrm{p}^2+\mathrm{q}^2-2 \mathrm{ph}+\mathrm{q}^2+\mathrm{k}^2-2 \mathrm{qk}\) \(4 \mathrm{q}=(\mathrm{h}-\mathrm{p})^2\) Hence, The locus of \((h, k)\) is \((x-p)^2=4 q y\)
119761
For how many values of \(k\), the line \(3 x-4 y=\lambda\) may touch the circle \(x^2+y^2-4 x-8 y-5=0\) ?
1 1
2 2
3 3
4 None of the values of \(\mathrm{k}\)
Explanation:
B The line \(3 \mathrm{x}-4 \mathrm{y}=\lambda \text { touches the circle }\) \(\quad \mathrm{x}^2+\mathrm{y}^2-4 \mathrm{x}-8 \mathrm{y}-5=0\) \(\quad \mathrm{C}(-\mathrm{g},-\mathrm{f})=(2,4)\) \(=\sqrt{\mathrm{g}^2+\mathrm{f}^2-\mathrm{c}}=\sqrt{4+16+5}=5\) \(\text { We know distance of a straight line } \mathrm{ax}+\mathrm{by}+\mathrm{c}=0\) \(\mathrm{ax}+\mathrm{by}+\mathrm{c}=0 \text { from any point }\left(\mathrm{x}_1, \mathrm{y}_1\right) \text { is }\) \(\mathrm{d}=\frac{\mathrm{ax} \mathrm{x}_1+\mathrm{by}_1+\mathrm{c}}{\sqrt{\mathrm{a}^2+\mathrm{b}^2}}\) \(5=\frac{\mathrm{ax}_1+\mathrm{by}_1+\mathrm{c}}{\sqrt{\mathrm{a}^2+\mathrm{b}^2}}=\frac{|6-16-\lambda|}{\sqrt{16+9}}=\frac{|-\lambda-10|}{5}\) \(\frac{-\lambda-10}{5}= \pm 5\) \(-\lambda-10= \pm 25= \pm 25-10=-35,15\) We know distance of a straight line \(a x+b y+c=0\) \(a x+b y+c=0\) from any point \(\left(x_1, y_1\right)\) is
SCRA-2015
Conic Section
119762
The line segment joining the foci of the hyperbola \(x^2-y^2+1=0\) is one of the diameters of a circle. The equation of the circle is
1 \(x^2+y^2=4\)
2 \(x^2+y^2=\sqrt{2}\)
3 \(x^2+y^2=2\)
4 \(x^2+y^2=2 \sqrt{2}\)
Explanation:
C Given, equation of hyperbola is \(x^2-y^2+1=0\) \(\Rightarrow \quad \mathrm{y}^2-\mathrm{x}^2=1\) On comparing it with \(\frac{y^2}{b^2}-\frac{x^2}{a^2}=1\). we get \(\quad \mathrm{a}=\mathrm{b}=1\) \(\text { now, } \mathrm{e}=\sqrt{1+\frac{\mathrm{a}^2}{\mathrm{~b}^2}}=\sqrt{1+\frac{1}{1}}=\sqrt{2}\) \(\therefore \text { foci }=(0, \pm \mathrm{be})=(0, \pm \sqrt{2})\) Since, line joining foci of hyperbola is diameter of circle. \(\therefore \text { center of circle }=\left(\frac{0+0}{2}, \frac{\sqrt{2}-\sqrt{2}}{2}\right)=(0,0)\) \(\text { and radius }=\frac{1}{2} \sqrt{(0-0)^2+(\sqrt{2}-(-\sqrt{2}))^2}\) \(\quad=\frac{1}{2} \sqrt{(2 \sqrt{2})^2}=\sqrt{2}\) \(\therefore\) Equation of circle will be \((\mathrm{x}-0)^2+(\mathrm{y}-0)^2=(\sqrt{2})^2\) \(\Rightarrow \quad \mathrm{x}^2+\mathrm{y}^2=2\)
WB JEE-2017
Conic Section
119763
If \(P(0,0), Q(1,0)\) and \(R\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)\) are three given points, then the centre of the circle for which the lines \(P Q, Q R\) and \(R P\) are the tangents is
119764
A variable circle passes through the fixed point \(A(p, q)\) and touches \(x\)-axis. The locus of the other end of the diameter through \(A\) is
1 \((x-p)^2=4 q y\)
2 \((x-q)^2=4 p y\)
3 \((\mathrm{y}-\mathrm{p})^2=4 \mathrm{qx}\)
4 \((y-q)^2=4 \mathrm{px}\)
Explanation:
A Let, ( \(\mathrm{h}, \mathrm{k}\) ) be the coordinates of the other end \(\mathrm{B}\) of the diameter through \(\mathrm{A}\). The coordinates of the centre are \(\left(\frac{\mathrm{p}+\mathrm{h}}{2}, \frac{\mathrm{q}+\mathrm{k}}{2}\right)\) Since, the circle touches \(\mathrm{x}\)-axis Therefore, \(\mid \mathrm{y} \text {-coordinates of its centre } \mid=\text { Radius }\) \(\frac{\mathrm{q}+\mathrm{k}}{2}=\frac{1}{2} \sqrt{(\mathrm{p}-\mathrm{h})^2+(\mathrm{q}-\mathrm{k})^2}\) \((\mathrm{q}+\mathrm{k})^2=(\mathrm{p}-\mathrm{h})^2+(\mathrm{q}-\mathrm{k})^2\) \(\mathrm{q}^2+\mathrm{k}^2+2 \mathrm{qk}=\mathrm{p}^2+\mathrm{q}^2-2 \mathrm{ph}+\mathrm{q}^2+\mathrm{k}^2-2 \mathrm{qk}\) \(4 \mathrm{q}=(\mathrm{h}-\mathrm{p})^2\) Hence, The locus of \((h, k)\) is \((x-p)^2=4 q y\)
