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Normal 0false false falseEN-US X-NONE X-NONE/* Style Definitions */table.MsoNormalTable{mso-style-name:Table Normal;mso-tstyle-rowband-size:0;mso-tstyle-colband-size:0;mso-style-noshow:yes;mso-style-priority:99;mso-style-parent:;mso-padding-alt:0in 5.4pt 0in 5.4pt;mso-para-margin-top:0in;mso-para-margin-right:0in;mso-para-margin-bottom:10.0pt;mso-para-margin-left:0in;line-height:115%;mso-pagination:widow-orphan;font-size:11.0pt;font-family:Calibri,sans-serif;mso-ascii-font-family:Calibri;mso-ascii-theme-font:minor-latin;mso-hansi-font-family:Calibri;mso-hansi-theme-font:minor-latin;mso-bidi-font-family:Times New Roman;mso-bidi-theme-font:minor-bidi;}4. A circle touches a g8iven straight line and cuts off a constant length 2d from another straight line perpendicular to the first straight line. The locus of the centre of the circle is(a) Hyperbola(b) Circle(c) Parabola(d) Pair of perpendicular lines[The two straight lines to be considered as coordinate axes]

Simran Bhatia , 11 Years ago
Grade 11
anser 1 Answers
Jitender Singh

Last Activity: 10 Years ago

Ans: (a) Hyperbola
Sol:
Let the centre of circle be (h, k), it touches the x-axis and cut a constant lenght of ‘2d’
on y-axis.
Equation of circle:
(x-h)^{2}+(y-k)^{2}=k^{2}
Circle intersection with y-axis (x = 0);
(-h)^{2}+(y-k)^{2}=k^{2}
y^{2}-2hk+h^{2}=0
y = \frac{k\pm \sqrt{k^{2}-4h^{2}}}{2}
Now, equate given length with difference of intersecting coordinates
\sqrt{k^{2}-4h^{2}} = 2d
{k^{2}-4h^{2}} = 4d^{2}
h\rightarrow x, k\rightarrow y
{y^{2}-4x^{2}} = 4d^{2}
Thanks & Regards
Jitender Singh
IIT Delhi
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