119899
The centres of those circles which touch the circle, \(x^2+y^2-8 x-8 y-4=0\), externally and also touch the \(\mathrm{X}\)-axis, lie on
1 a circle
2 an ellipse which is not a circle
3 a hyperbola
4 a parabola
Explanation:
D Given circle equation. \(\mathrm{C}_1=\mathrm{x}^2+\mathrm{y}^2-8 \mathrm{x}-8 \mathrm{y}-4=0\) Let a circle \(c_2 \Rightarrow(x-h)^2+(y-k)^2=r^2\) touch the given circle. \(\therefore \mathrm{r}_1=\sqrt{\mathrm{g}^2+\mathrm{f}^2-\mathrm{c}}=\sqrt{16+16+4}=\sqrt{36}\) \(\mathrm{r}_1=6\) \(\mathrm{r}_2=\mathrm{r}\) \((\mathrm{h}-4)^2+(\mathrm{k}-4)^2=(6+\mathrm{r})^2\) \(\mathrm{c}_2 \text { touches } \mathrm{x}-\mathrm{axis} \mathrm{so,} \mathrm{r}=\mathrm{r}\) \((\mathrm{h}-4)^2+(\mathrm{k}-4)^2=(6+\mathrm{k})^2\) \(\mathrm{~h}^2+16-8 \mathrm{~h}+\mathrm{k}^2+16-8 \mathrm{k}=36+\mathrm{k}^2+12 \mathrm{k}\) \(\left(\mathrm{h}^2+16-8 \mathrm{~h}\right)-20 \mathrm{k}-10=0\) \((\mathrm{~h}-4)^2-20 \mathrm{k}-20=0\) \((\mathrm{~h}-4)^2=20 \mathrm{k}+20\) \(\text { Replacing (h, k) by (x, y) }\) \((\mathrm{x}-4)^2=4(5 \mathrm{y}+5)\) So, locus is a parabola
JEE Main-2016
Conic Section
119900
If the line \(a x+y=c\), touches both the curves \(x^2\) \(+y^2=1\) and \(y^2=4 \sqrt{2} x\), then \(|c|\) is equal to
1 \(\frac{1}{\sqrt{2}}\)
2 2
3 \(\sqrt{2}\)
4 \(\frac{1}{2}\)
Explanation:
C Given line equation \(\text { ax }+\mathrm{y}=\mathrm{c}\) \(\text { Tangent to } \mathrm{y}^2=4 \sqrt{2} \mathrm{x} \text { is } \mathrm{y}=\mathrm{mx}+\frac{\sqrt{2}}{\mathrm{~m}} \text { is also tangent }\) \(\text { to } \mathrm{x}^2+\mathrm{y}^2=1\) \(\left|\frac{\sqrt{2} / \mathrm{m}}{\sqrt{1+\mathrm{m}^2}}\right|=1\) \(\mathrm{~m}= \pm 1\) \(\text { The eq } \mathrm{q}^{\mathrm{n}} \text {. of tangent will be } \mathrm{y}=\mathrm{x}+\sqrt{2}\) \(\text { or } \mathrm{y}=-\mathrm{x}-\sqrt{2}\) \(\text { Compare the above eq } \mathrm{e}^{\mathrm{n}} \text {. with } \mathrm{y}=-\mathrm{ax}+\mathrm{c}\) \(\therefore \mathrm{a}= \pm 1 \text { and } \mathrm{c}= \pm \sqrt{2}\) \(\therefore \quad|\mathrm{c}|=\sqrt{2}\) Tangent to \(y^2=4 \sqrt{2} x\) is \(y=m x+\frac{\sqrt{2}}{m}\) is also tangent to \(\mathrm{x}^2+\mathrm{y}^2=1\) The eq \(\mathrm{q}^{\mathrm{n}}\). of tangent will be \(\mathrm{y}=\mathrm{x}+\sqrt{2}\) or \(\mathrm{y}=-\mathrm{x}-\sqrt{2}\) Compare the above eq \({ }^n\). with \(\mathrm{y}=-\mathrm{ax}+\mathrm{c}\)
JEE Main 10.04.2019
Conic Section
119901
A tangent is drawn to the parabola \(y^2=6 x\), which is perpendicular to the line \(2 x+y=1\). Which of the following points does not lie on it?
1 \((-6,0)\)
2 \((4,5)\)
3 \((5,4)\)
4 \((0,3)\)
Explanation:
C Given equation of parabola, \(\mathrm{y}^2=6 \mathrm{x} \text { and line }\) \(\text { equation, } 2 \mathrm{x}+\mathrm{y}=1\) \(\text { Since, the above line is perpendicular to tangent }\) \(\therefore \quad \text { slope of tangent }\) \(\mathrm{m}(-2)=-1\) \(\mathrm{~m}=\frac{1}{2}\) \(\text { Equation } \mathrm{y}=\mathrm{mx}+\frac{\mathrm{a}}{\mathrm{m}}\) \(\qquad \mathrm{y}=\frac{1}{2} \mathrm{x}+\frac{3}{2 \times \frac{1}{2}} \quad\left(\mathrm{a}=\frac{6}{4}=\frac{3}{2}\right)\) \(\quad \mathrm{y}=\frac{\mathrm{x}}{2}+3\) \(2 \mathrm{y}=\mathrm{x}+6\) \(\text { Hence, the point }(5,4) \text { does not lie on it. }\)
JEE Main 25.02.2021
Conic Section
119903
If the two circles \((x-1)^2+(y-3)^2=\mathbf{r}^2\) and \(x^2+\) \(y^2-8 x+2 y+8=0\) intersect in two distinct points, then
1 \(2\lt \) r \(\lt 8\)
2 \(\mathrm{r}\lt 2\)
3 \(r=2\)
4 \(r>2\) [-2002]
Explanation:
A Here we have two circles (i) \((x-1)^2+(y-3)^2=r^2\) and \(x^2+y^2-8 x+2 y+8=0\) \(\text { (ii) }(x-4)^2+(y+1)^2=9\)Coordinates of the centre of the two circles or \((1,3)\) and \((4,-1)\) respectively Distance between two center \(\mathrm{d}=\sqrt{(1-4)^2+(3+1)^2}=\sqrt{9+16}=5\) As the two circles intersects at two distinct point \(\therefore \mathrm{d}>\left|\mathrm{r}_{\mathrm{A}}-\mathrm{r}_{\mathrm{B}}\right| \text { and } \mathrm{d}\lt \left|\mathrm{r}_{\mathrm{A}}+\mathrm{r}_{\mathrm{B}}\right|\) \(\mathrm{d}=5, \mathrm{r}_{\mathrm{A}}=\mathrm{r}, \mathrm{r}_{\mathrm{B}}=3\) So we get- \(5>\mathrm{r}-3 \Rightarrow \mathrm{r}\lt 8\) \(5>\mathrm{r}+3 \Rightarrow \mathrm{r}>2\) \(\text { So, } 2\lt \mathrm{r}\lt 8\)
119899
The centres of those circles which touch the circle, \(x^2+y^2-8 x-8 y-4=0\), externally and also touch the \(\mathrm{X}\)-axis, lie on
1 a circle
2 an ellipse which is not a circle
3 a hyperbola
4 a parabola
Explanation:
D Given circle equation. \(\mathrm{C}_1=\mathrm{x}^2+\mathrm{y}^2-8 \mathrm{x}-8 \mathrm{y}-4=0\) Let a circle \(c_2 \Rightarrow(x-h)^2+(y-k)^2=r^2\) touch the given circle. \(\therefore \mathrm{r}_1=\sqrt{\mathrm{g}^2+\mathrm{f}^2-\mathrm{c}}=\sqrt{16+16+4}=\sqrt{36}\) \(\mathrm{r}_1=6\) \(\mathrm{r}_2=\mathrm{r}\) \((\mathrm{h}-4)^2+(\mathrm{k}-4)^2=(6+\mathrm{r})^2\) \(\mathrm{c}_2 \text { touches } \mathrm{x}-\mathrm{axis} \mathrm{so,} \mathrm{r}=\mathrm{r}\) \((\mathrm{h}-4)^2+(\mathrm{k}-4)^2=(6+\mathrm{k})^2\) \(\mathrm{~h}^2+16-8 \mathrm{~h}+\mathrm{k}^2+16-8 \mathrm{k}=36+\mathrm{k}^2+12 \mathrm{k}\) \(\left(\mathrm{h}^2+16-8 \mathrm{~h}\right)-20 \mathrm{k}-10=0\) \((\mathrm{~h}-4)^2-20 \mathrm{k}-20=0\) \((\mathrm{~h}-4)^2=20 \mathrm{k}+20\) \(\text { Replacing (h, k) by (x, y) }\) \((\mathrm{x}-4)^2=4(5 \mathrm{y}+5)\) So, locus is a parabola
JEE Main-2016
Conic Section
119900
If the line \(a x+y=c\), touches both the curves \(x^2\) \(+y^2=1\) and \(y^2=4 \sqrt{2} x\), then \(|c|\) is equal to
1 \(\frac{1}{\sqrt{2}}\)
2 2
3 \(\sqrt{2}\)
4 \(\frac{1}{2}\)
Explanation:
C Given line equation \(\text { ax }+\mathrm{y}=\mathrm{c}\) \(\text { Tangent to } \mathrm{y}^2=4 \sqrt{2} \mathrm{x} \text { is } \mathrm{y}=\mathrm{mx}+\frac{\sqrt{2}}{\mathrm{~m}} \text { is also tangent }\) \(\text { to } \mathrm{x}^2+\mathrm{y}^2=1\) \(\left|\frac{\sqrt{2} / \mathrm{m}}{\sqrt{1+\mathrm{m}^2}}\right|=1\) \(\mathrm{~m}= \pm 1\) \(\text { The eq } \mathrm{q}^{\mathrm{n}} \text {. of tangent will be } \mathrm{y}=\mathrm{x}+\sqrt{2}\) \(\text { or } \mathrm{y}=-\mathrm{x}-\sqrt{2}\) \(\text { Compare the above eq } \mathrm{e}^{\mathrm{n}} \text {. with } \mathrm{y}=-\mathrm{ax}+\mathrm{c}\) \(\therefore \mathrm{a}= \pm 1 \text { and } \mathrm{c}= \pm \sqrt{2}\) \(\therefore \quad|\mathrm{c}|=\sqrt{2}\) Tangent to \(y^2=4 \sqrt{2} x\) is \(y=m x+\frac{\sqrt{2}}{m}\) is also tangent to \(\mathrm{x}^2+\mathrm{y}^2=1\) The eq \(\mathrm{q}^{\mathrm{n}}\). of tangent will be \(\mathrm{y}=\mathrm{x}+\sqrt{2}\) or \(\mathrm{y}=-\mathrm{x}-\sqrt{2}\) Compare the above eq \({ }^n\). with \(\mathrm{y}=-\mathrm{ax}+\mathrm{c}\)
JEE Main 10.04.2019
Conic Section
119901
A tangent is drawn to the parabola \(y^2=6 x\), which is perpendicular to the line \(2 x+y=1\). Which of the following points does not lie on it?
1 \((-6,0)\)
2 \((4,5)\)
3 \((5,4)\)
4 \((0,3)\)
Explanation:
C Given equation of parabola, \(\mathrm{y}^2=6 \mathrm{x} \text { and line }\) \(\text { equation, } 2 \mathrm{x}+\mathrm{y}=1\) \(\text { Since, the above line is perpendicular to tangent }\) \(\therefore \quad \text { slope of tangent }\) \(\mathrm{m}(-2)=-1\) \(\mathrm{~m}=\frac{1}{2}\) \(\text { Equation } \mathrm{y}=\mathrm{mx}+\frac{\mathrm{a}}{\mathrm{m}}\) \(\qquad \mathrm{y}=\frac{1}{2} \mathrm{x}+\frac{3}{2 \times \frac{1}{2}} \quad\left(\mathrm{a}=\frac{6}{4}=\frac{3}{2}\right)\) \(\quad \mathrm{y}=\frac{\mathrm{x}}{2}+3\) \(2 \mathrm{y}=\mathrm{x}+6\) \(\text { Hence, the point }(5,4) \text { does not lie on it. }\)
JEE Main 25.02.2021
Conic Section
119903
If the two circles \((x-1)^2+(y-3)^2=\mathbf{r}^2\) and \(x^2+\) \(y^2-8 x+2 y+8=0\) intersect in two distinct points, then
1 \(2\lt \) r \(\lt 8\)
2 \(\mathrm{r}\lt 2\)
3 \(r=2\)
4 \(r>2\) [-2002]
Explanation:
A Here we have two circles (i) \((x-1)^2+(y-3)^2=r^2\) and \(x^2+y^2-8 x+2 y+8=0\) \(\text { (ii) }(x-4)^2+(y+1)^2=9\)Coordinates of the centre of the two circles or \((1,3)\) and \((4,-1)\) respectively Distance between two center \(\mathrm{d}=\sqrt{(1-4)^2+(3+1)^2}=\sqrt{9+16}=5\) As the two circles intersects at two distinct point \(\therefore \mathrm{d}>\left|\mathrm{r}_{\mathrm{A}}-\mathrm{r}_{\mathrm{B}}\right| \text { and } \mathrm{d}\lt \left|\mathrm{r}_{\mathrm{A}}+\mathrm{r}_{\mathrm{B}}\right|\) \(\mathrm{d}=5, \mathrm{r}_{\mathrm{A}}=\mathrm{r}, \mathrm{r}_{\mathrm{B}}=3\) So we get- \(5>\mathrm{r}-3 \Rightarrow \mathrm{r}\lt 8\) \(5>\mathrm{r}+3 \Rightarrow \mathrm{r}>2\) \(\text { So, } 2\lt \mathrm{r}\lt 8\)
119899
The centres of those circles which touch the circle, \(x^2+y^2-8 x-8 y-4=0\), externally and also touch the \(\mathrm{X}\)-axis, lie on
1 a circle
2 an ellipse which is not a circle
3 a hyperbola
4 a parabola
Explanation:
D Given circle equation. \(\mathrm{C}_1=\mathrm{x}^2+\mathrm{y}^2-8 \mathrm{x}-8 \mathrm{y}-4=0\) Let a circle \(c_2 \Rightarrow(x-h)^2+(y-k)^2=r^2\) touch the given circle. \(\therefore \mathrm{r}_1=\sqrt{\mathrm{g}^2+\mathrm{f}^2-\mathrm{c}}=\sqrt{16+16+4}=\sqrt{36}\) \(\mathrm{r}_1=6\) \(\mathrm{r}_2=\mathrm{r}\) \((\mathrm{h}-4)^2+(\mathrm{k}-4)^2=(6+\mathrm{r})^2\) \(\mathrm{c}_2 \text { touches } \mathrm{x}-\mathrm{axis} \mathrm{so,} \mathrm{r}=\mathrm{r}\) \((\mathrm{h}-4)^2+(\mathrm{k}-4)^2=(6+\mathrm{k})^2\) \(\mathrm{~h}^2+16-8 \mathrm{~h}+\mathrm{k}^2+16-8 \mathrm{k}=36+\mathrm{k}^2+12 \mathrm{k}\) \(\left(\mathrm{h}^2+16-8 \mathrm{~h}\right)-20 \mathrm{k}-10=0\) \((\mathrm{~h}-4)^2-20 \mathrm{k}-20=0\) \((\mathrm{~h}-4)^2=20 \mathrm{k}+20\) \(\text { Replacing (h, k) by (x, y) }\) \((\mathrm{x}-4)^2=4(5 \mathrm{y}+5)\) So, locus is a parabola
JEE Main-2016
Conic Section
119900
If the line \(a x+y=c\), touches both the curves \(x^2\) \(+y^2=1\) and \(y^2=4 \sqrt{2} x\), then \(|c|\) is equal to
1 \(\frac{1}{\sqrt{2}}\)
2 2
3 \(\sqrt{2}\)
4 \(\frac{1}{2}\)
Explanation:
C Given line equation \(\text { ax }+\mathrm{y}=\mathrm{c}\) \(\text { Tangent to } \mathrm{y}^2=4 \sqrt{2} \mathrm{x} \text { is } \mathrm{y}=\mathrm{mx}+\frac{\sqrt{2}}{\mathrm{~m}} \text { is also tangent }\) \(\text { to } \mathrm{x}^2+\mathrm{y}^2=1\) \(\left|\frac{\sqrt{2} / \mathrm{m}}{\sqrt{1+\mathrm{m}^2}}\right|=1\) \(\mathrm{~m}= \pm 1\) \(\text { The eq } \mathrm{q}^{\mathrm{n}} \text {. of tangent will be } \mathrm{y}=\mathrm{x}+\sqrt{2}\) \(\text { or } \mathrm{y}=-\mathrm{x}-\sqrt{2}\) \(\text { Compare the above eq } \mathrm{e}^{\mathrm{n}} \text {. with } \mathrm{y}=-\mathrm{ax}+\mathrm{c}\) \(\therefore \mathrm{a}= \pm 1 \text { and } \mathrm{c}= \pm \sqrt{2}\) \(\therefore \quad|\mathrm{c}|=\sqrt{2}\) Tangent to \(y^2=4 \sqrt{2} x\) is \(y=m x+\frac{\sqrt{2}}{m}\) is also tangent to \(\mathrm{x}^2+\mathrm{y}^2=1\) The eq \(\mathrm{q}^{\mathrm{n}}\). of tangent will be \(\mathrm{y}=\mathrm{x}+\sqrt{2}\) or \(\mathrm{y}=-\mathrm{x}-\sqrt{2}\) Compare the above eq \({ }^n\). with \(\mathrm{y}=-\mathrm{ax}+\mathrm{c}\)
JEE Main 10.04.2019
Conic Section
119901
A tangent is drawn to the parabola \(y^2=6 x\), which is perpendicular to the line \(2 x+y=1\). Which of the following points does not lie on it?
1 \((-6,0)\)
2 \((4,5)\)
3 \((5,4)\)
4 \((0,3)\)
Explanation:
C Given equation of parabola, \(\mathrm{y}^2=6 \mathrm{x} \text { and line }\) \(\text { equation, } 2 \mathrm{x}+\mathrm{y}=1\) \(\text { Since, the above line is perpendicular to tangent }\) \(\therefore \quad \text { slope of tangent }\) \(\mathrm{m}(-2)=-1\) \(\mathrm{~m}=\frac{1}{2}\) \(\text { Equation } \mathrm{y}=\mathrm{mx}+\frac{\mathrm{a}}{\mathrm{m}}\) \(\qquad \mathrm{y}=\frac{1}{2} \mathrm{x}+\frac{3}{2 \times \frac{1}{2}} \quad\left(\mathrm{a}=\frac{6}{4}=\frac{3}{2}\right)\) \(\quad \mathrm{y}=\frac{\mathrm{x}}{2}+3\) \(2 \mathrm{y}=\mathrm{x}+6\) \(\text { Hence, the point }(5,4) \text { does not lie on it. }\)
JEE Main 25.02.2021
Conic Section
119903
If the two circles \((x-1)^2+(y-3)^2=\mathbf{r}^2\) and \(x^2+\) \(y^2-8 x+2 y+8=0\) intersect in two distinct points, then
1 \(2\lt \) r \(\lt 8\)
2 \(\mathrm{r}\lt 2\)
3 \(r=2\)
4 \(r>2\) [-2002]
Explanation:
A Here we have two circles (i) \((x-1)^2+(y-3)^2=r^2\) and \(x^2+y^2-8 x+2 y+8=0\) \(\text { (ii) }(x-4)^2+(y+1)^2=9\)Coordinates of the centre of the two circles or \((1,3)\) and \((4,-1)\) respectively Distance between two center \(\mathrm{d}=\sqrt{(1-4)^2+(3+1)^2}=\sqrt{9+16}=5\) As the two circles intersects at two distinct point \(\therefore \mathrm{d}>\left|\mathrm{r}_{\mathrm{A}}-\mathrm{r}_{\mathrm{B}}\right| \text { and } \mathrm{d}\lt \left|\mathrm{r}_{\mathrm{A}}+\mathrm{r}_{\mathrm{B}}\right|\) \(\mathrm{d}=5, \mathrm{r}_{\mathrm{A}}=\mathrm{r}, \mathrm{r}_{\mathrm{B}}=3\) So we get- \(5>\mathrm{r}-3 \Rightarrow \mathrm{r}\lt 8\) \(5>\mathrm{r}+3 \Rightarrow \mathrm{r}>2\) \(\text { So, } 2\lt \mathrm{r}\lt 8\)
119899
The centres of those circles which touch the circle, \(x^2+y^2-8 x-8 y-4=0\), externally and also touch the \(\mathrm{X}\)-axis, lie on
1 a circle
2 an ellipse which is not a circle
3 a hyperbola
4 a parabola
Explanation:
D Given circle equation. \(\mathrm{C}_1=\mathrm{x}^2+\mathrm{y}^2-8 \mathrm{x}-8 \mathrm{y}-4=0\) Let a circle \(c_2 \Rightarrow(x-h)^2+(y-k)^2=r^2\) touch the given circle. \(\therefore \mathrm{r}_1=\sqrt{\mathrm{g}^2+\mathrm{f}^2-\mathrm{c}}=\sqrt{16+16+4}=\sqrt{36}\) \(\mathrm{r}_1=6\) \(\mathrm{r}_2=\mathrm{r}\) \((\mathrm{h}-4)^2+(\mathrm{k}-4)^2=(6+\mathrm{r})^2\) \(\mathrm{c}_2 \text { touches } \mathrm{x}-\mathrm{axis} \mathrm{so,} \mathrm{r}=\mathrm{r}\) \((\mathrm{h}-4)^2+(\mathrm{k}-4)^2=(6+\mathrm{k})^2\) \(\mathrm{~h}^2+16-8 \mathrm{~h}+\mathrm{k}^2+16-8 \mathrm{k}=36+\mathrm{k}^2+12 \mathrm{k}\) \(\left(\mathrm{h}^2+16-8 \mathrm{~h}\right)-20 \mathrm{k}-10=0\) \((\mathrm{~h}-4)^2-20 \mathrm{k}-20=0\) \((\mathrm{~h}-4)^2=20 \mathrm{k}+20\) \(\text { Replacing (h, k) by (x, y) }\) \((\mathrm{x}-4)^2=4(5 \mathrm{y}+5)\) So, locus is a parabola
JEE Main-2016
Conic Section
119900
If the line \(a x+y=c\), touches both the curves \(x^2\) \(+y^2=1\) and \(y^2=4 \sqrt{2} x\), then \(|c|\) is equal to
1 \(\frac{1}{\sqrt{2}}\)
2 2
3 \(\sqrt{2}\)
4 \(\frac{1}{2}\)
Explanation:
C Given line equation \(\text { ax }+\mathrm{y}=\mathrm{c}\) \(\text { Tangent to } \mathrm{y}^2=4 \sqrt{2} \mathrm{x} \text { is } \mathrm{y}=\mathrm{mx}+\frac{\sqrt{2}}{\mathrm{~m}} \text { is also tangent }\) \(\text { to } \mathrm{x}^2+\mathrm{y}^2=1\) \(\left|\frac{\sqrt{2} / \mathrm{m}}{\sqrt{1+\mathrm{m}^2}}\right|=1\) \(\mathrm{~m}= \pm 1\) \(\text { The eq } \mathrm{q}^{\mathrm{n}} \text {. of tangent will be } \mathrm{y}=\mathrm{x}+\sqrt{2}\) \(\text { or } \mathrm{y}=-\mathrm{x}-\sqrt{2}\) \(\text { Compare the above eq } \mathrm{e}^{\mathrm{n}} \text {. with } \mathrm{y}=-\mathrm{ax}+\mathrm{c}\) \(\therefore \mathrm{a}= \pm 1 \text { and } \mathrm{c}= \pm \sqrt{2}\) \(\therefore \quad|\mathrm{c}|=\sqrt{2}\) Tangent to \(y^2=4 \sqrt{2} x\) is \(y=m x+\frac{\sqrt{2}}{m}\) is also tangent to \(\mathrm{x}^2+\mathrm{y}^2=1\) The eq \(\mathrm{q}^{\mathrm{n}}\). of tangent will be \(\mathrm{y}=\mathrm{x}+\sqrt{2}\) or \(\mathrm{y}=-\mathrm{x}-\sqrt{2}\) Compare the above eq \({ }^n\). with \(\mathrm{y}=-\mathrm{ax}+\mathrm{c}\)
JEE Main 10.04.2019
Conic Section
119901
A tangent is drawn to the parabola \(y^2=6 x\), which is perpendicular to the line \(2 x+y=1\). Which of the following points does not lie on it?
1 \((-6,0)\)
2 \((4,5)\)
3 \((5,4)\)
4 \((0,3)\)
Explanation:
C Given equation of parabola, \(\mathrm{y}^2=6 \mathrm{x} \text { and line }\) \(\text { equation, } 2 \mathrm{x}+\mathrm{y}=1\) \(\text { Since, the above line is perpendicular to tangent }\) \(\therefore \quad \text { slope of tangent }\) \(\mathrm{m}(-2)=-1\) \(\mathrm{~m}=\frac{1}{2}\) \(\text { Equation } \mathrm{y}=\mathrm{mx}+\frac{\mathrm{a}}{\mathrm{m}}\) \(\qquad \mathrm{y}=\frac{1}{2} \mathrm{x}+\frac{3}{2 \times \frac{1}{2}} \quad\left(\mathrm{a}=\frac{6}{4}=\frac{3}{2}\right)\) \(\quad \mathrm{y}=\frac{\mathrm{x}}{2}+3\) \(2 \mathrm{y}=\mathrm{x}+6\) \(\text { Hence, the point }(5,4) \text { does not lie on it. }\)
JEE Main 25.02.2021
Conic Section
119903
If the two circles \((x-1)^2+(y-3)^2=\mathbf{r}^2\) and \(x^2+\) \(y^2-8 x+2 y+8=0\) intersect in two distinct points, then
1 \(2\lt \) r \(\lt 8\)
2 \(\mathrm{r}\lt 2\)
3 \(r=2\)
4 \(r>2\) [-2002]
Explanation:
A Here we have two circles (i) \((x-1)^2+(y-3)^2=r^2\) and \(x^2+y^2-8 x+2 y+8=0\) \(\text { (ii) }(x-4)^2+(y+1)^2=9\)Coordinates of the centre of the two circles or \((1,3)\) and \((4,-1)\) respectively Distance between two center \(\mathrm{d}=\sqrt{(1-4)^2+(3+1)^2}=\sqrt{9+16}=5\) As the two circles intersects at two distinct point \(\therefore \mathrm{d}>\left|\mathrm{r}_{\mathrm{A}}-\mathrm{r}_{\mathrm{B}}\right| \text { and } \mathrm{d}\lt \left|\mathrm{r}_{\mathrm{A}}+\mathrm{r}_{\mathrm{B}}\right|\) \(\mathrm{d}=5, \mathrm{r}_{\mathrm{A}}=\mathrm{r}, \mathrm{r}_{\mathrm{B}}=3\) So we get- \(5>\mathrm{r}-3 \Rightarrow \mathrm{r}\lt 8\) \(5>\mathrm{r}+3 \Rightarrow \mathrm{r}>2\) \(\text { So, } 2\lt \mathrm{r}\lt 8\)
